A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3 \mathrm{~m} / \mathrm{s}$ (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm.
Explanation:
Total initial kinetic energy
$ =\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \mathrm{I} \omega^2 $
$\mathrm{v}=\mathrm{R} \omega$ (for pure rolling)
$ \mathrm{K} . E C=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \times \frac{2}{3} m \mathrm{R}^2 \times \frac{\mathrm{v}^2}{\mathrm{R}^2}=\frac{5}{6} m \mathrm{v}^2 $
Energy remains conserve during whole journey.
$\mathrm{K}_{\cdot} \mathrm{E}_i+\mathrm{P.E}_{\cdot i}=\mathrm{K}_{\cdot} \mathrm{E}_f+\mathrm{P.E}_{\cdot f}$
$ \begin{aligned} & \Rightarrow \frac{5}{2} m \mathrm{v}^2=m g \mathrm{H} ~~~~~~~(\because {K.E.}_f=0)\\\\ & \Rightarrow \mathrm{H}=\frac{5}{6} \times \frac{\mathrm{v}^2}{g} \\\\ & =\frac{5 \times(3)^2}{6 \times 10} \\\\ & =\frac{15}{20} \mathrm{~m}=0.75 \mathrm{~m}=75 \mathrm{~cm} \end{aligned} $
The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is $\frac{1}{x} \mathrm{MR}^{2}$, where $\mathrm{R}$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be __________.
Explanation:
To solve this problem, we need to understand the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass distribution of the object and the axis of rotation.
For a continuous object like a semicircular ring, we can calculate the moment of inertia by integrating over the entire object. Here's how we can approach this problem:
1. Divide the semicircular ring into small mass elements: Imagine the semicircular ring divided into infinitesimally small mass elements, each with mass $dm$.
2. Calculate the moment of inertia of each element: The moment of inertia of each element about the axis passing through the center and perpendicular to the plane of the ring is given by $dI = dmR^2$, where R is the radius of the ring.
3. Integrate to find the total moment of inertia: To find the total moment of inertia, we need to integrate $dI$ over the entire ring. This means integrating from $0$ to $\pi$ (the angle spanned by the semicircle) with respect to the angle $\theta$.
4. Relate $dm$ to the total mass: Since the ring has a uniform mass distribution, we can express the mass of each element $dm$ as a fraction of the total mass $M$: $dm = \frac{M}{πR} Rd\theta = \frac{M}{\pi} d\theta$.
Now, let's perform the integration:
$I = \int_{0}^{\pi} dI = \int_{0}^{\pi} dmR^2 = \int_{0}^{\pi} \frac{M}{\pi} d\theta R^2$
$I = \frac{MR^2}{\pi} \int_{0}^{\pi} d\theta = \frac{MR^2}{\pi} [\theta]_{0}^{\pi}$
$I = \frac{MR^2}{\pi} [\pi - 0] = MR^2$
Therefore, the moment of inertia of the semicircular ring about the given axis is $MR^2$. Comparing this to the given formula, we find that $x = \boxed{1}$.
A ring and a solid sphere rotating about an axis passing through their centers have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is $\sqrt{\frac{2}{x}}$. The value of $x$ is ___________.
Explanation:
Given that the radii of gyration for the ring and the solid sphere are equal, we have:
$ K_1 = K_2 $
For the ring, the moment of inertia is:
$ I_{ring} = mR_1^2 = mK_1^2 $
Thus, the radius of gyration for the ring is:
$ K_1 = R_1 $
For the solid sphere, the moment of inertia is:
$ I_{sphere} = \frac{2}{5}m'R_2^2 = m'K_2^2 $
Hence, the radius of gyration for the solid sphere is:
$ K_2 = \sqrt{\frac{2}{5}}R_2 $
Since the radii of gyration are equal:
$ R_1 = \sqrt{\frac{2}{5}}R_2 $
Therefore, the ratio of the radius of the ring to that of the sphere is:
$ \frac{R_1}{R_2} = \sqrt{\frac{2}{5}} $
So, the value of $x$ is:
$ x = 5 $
Two identical solid spheres each of mass $2 \mathrm{~kg}$ and radii $10 \mathrm{~cm}$ are fixed at the ends of a light rod. The separation between the centres of the spheres is $40 \mathrm{~cm}$. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ $\times 10^{-3} \mathrm{~kg}~\mathrm{m}^{2}$
Explanation:
The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.
First, let’s identify the moment of inertia of each solid sphere about its own center, which is given by the formula:
$I_{\text{sphere}} = \frac{2}{5} m r^2$
Where:
- $m$ is the mass of the sphere = $2 \mathrm{~kg}$
- $r$ is the radius of the sphere = $0.1 \mathrm{~m}$
Substituting the values:
$I_{\text{sphere}} = \frac{2}{5} \times 2 \mathrm{~kg} \times (0.1 \mathrm{~m})^2 = \frac{4}{5} \times 0.01 \mathrm{~kg}~\mathrm{m}^2 = 0.008 \mathrm{~kg}~\mathrm{m}^2$
Now, we need the moment of inertia of the two spheres about the axis passing through the midpoint of the rod. This requires using the parallel axis theorem, which states:
$I_{\text{total}} = I_{\text{sphere}} + m d^2$
Where:
- $d$ is the distance from the center of the sphere to the axis through the rod’s midpoint = $0.2 \mathrm{~m}$
Calculating the additional inertia due to the parallel axis theorem for one sphere:
$I_{\text{parallel}} = m d^2 = 2 \mathrm{~kg} \times (0.2 \mathrm{~m})^2 = 2 \mathrm{~kg} \times 0.04 \mathrm{~m}^2 = 0.08 \mathrm{~kg}~\mathrm{m}^2$
The total moment of inertia for one sphere about the midpoint of the rod is:
$I_{\text{one sphere, total}} = I_{\text{sphere}} + I_{\text{parallel}} = 0.008 \mathrm{~kg}~\mathrm{m}^2 + 0.08 \mathrm{~kg}~\mathrm{m}^2 = 0.088 \mathrm{~kg}~\mathrm{m}^2$
Since there are two identical spheres, the total moment of inertia of the system is:
$I_{\text{system}} = 2 \times 0.088 \mathrm{~kg}~\mathrm{m}^2 = 0.176 \mathrm{~kg}~\mathrm{m}^2$
Converting the result to the given form:
$0.176 \mathrm{~kg}~\mathrm{m}^2 = 176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$
So, the moment of inertia of the system about the given axis is:
$176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$
Moment of inertia of a disc of mass '$M$' and radius '$R$' about any of its diameter is $\frac{M R^{2}}{4}$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\frac{x}{2}$ MR$^{2}$. The value of $x$ is ___________.
Explanation:
$\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 \\\\ & =\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2 \\\\ & =\frac{3}{2} \mathrm{MR}^2 \\\\ & \mathrm{x}=3\end{aligned}$
A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \mathrm{~cm}$. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is __________ $\mathrm{ms}^{-1}$. (Given $\mathrm{g}=10 \mathrm{~ms}^{-2}$)
Explanation:
$\Rightarrow m g\left[\frac{30}{100}\right]=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m v^{2}}{2}$
$\Rightarrow 0.3 \times 10=\frac{3}{4} v^{2}$
$\Rightarrow v^{2}=4$
$\Rightarrow v=2 \mathrm{~m} / \mathrm{s}$
Explanation:
$ \begin{aligned} & \text { so } R^{2}=\frac{m}{\rho \pi t} \\\\ & I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t} \end{aligned} $
So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$
So $x=5$
A solid sphere of mass $1 \mathrm{~kg}$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3} \mathrm{~J}$. The speed of the centre of mass of the sphere is __________ $\operatorname{cm~s}^{-1}$
Explanation:
$ \Rightarrow $ $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{MR}^{2}\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=7 \times 10^{-3}$
$ \Rightarrow $ $\frac{1}{2} \mathrm{MV}^{2}\left[1+\frac{2}{5}\right]=7 \times 10^{-3}$
$ \Rightarrow $ $\frac{1}{2}(1)\left(\mathrm{V}^{2}\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}$
$ \Rightarrow $ $\mathrm{V}^{2}=10^{-2}$
$ \Rightarrow $ $\mathrm{V}=10^{-1}=0.1 \mathrm{~m} / \mathrm{s}=10 \mathrm{~cm} / \mathrm{s}$
A uniform disc of mass $0.5 \mathrm{~kg}$ and radius $r$ is projected with velocity $18 \mathrm{~m} / \mathrm{s}$ at $\mathrm{t}=0$ s on a rough horizontal surface. It starts off with a purely sliding motion at $\mathrm{t}=0 \mathrm{~s}$. After $2 \mathrm{~s}$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2 \mathrm{~s}$ will be __________ $\mathrm{J}$ (given, coefficient of friction is $0.3$ and $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ).
Explanation:
$v = {v_0} - \mu gt$
$ \Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12$ m/s
$\Rightarrow$ Kinetic energy $ = {1 \over 2}m{v^2} + {1 \over 2}{{m{v^2}} \over 2}$
$ = {3 \over 4}m{v^2} = {3 \over 4} \times 0.5 \times 144\,\mathrm{J} = 54\,\mathrm{J}$
A thin uniform rod of length $2 \mathrm{~m}$, cross sectional area '$A$' and density '$\mathrm{d}$' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$. If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\frac{\alpha E}{A d}}$ then value of $\alpha$ is ______________.
Explanation:
Kinetic energy of rod $E = {1 \over 2}{{m{l^2}} \over {12}}{\omega ^2}$
or $\omega = \sqrt {{{24E} \over {m{l^2}}}} = \sqrt {{{24E} \over {d \times A \times {l^3}}}} $
$ \Rightarrow \omega = \sqrt {{{24E} \over {dA{2^3}}}} $
$ = \sqrt {{{3E} \over {Ad}}} $
So, $\alpha = 3$
A particle of mass 100 g is projected at time t = 0 with a speed 20 ms$^{-1}$ at an angle 45$^\circ$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t = 2s is found to be $\mathrm{\sqrt K~kg~m^2/s}$. The value of K is ___________.
(Take g = 10 ms$^{-2}$)
Explanation:
Horizontal displacement $x = v\cos \theta t$
$ = 10\sqrt 2 t$
So torque of weight about point of projection is
$\tau = mgx\,.\,( - \widehat k)$
${{d\overrightarrow L } \over {dt}} = mgx( - \widehat k)$
$\int_0^{\overrightarrow L } {d\overrightarrow L = 0.1 \times 10 \times 10\sqrt 2 \int_0^2 {t\,dt( - \widehat k)} } $
$\overrightarrow L = - 20\sqrt 2 \widehat k$
$|\overrightarrow L | = 20\sqrt 2 = \sqrt {800} $ kg m$^2$/s
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be _______ ms$^{-1}$.
Explanation:
$ \begin{aligned} & \frac{7}{10} m v_{\mathrm{cm}}^{2}=2240 \\\\ & v_{\mathrm{cm}}=\sqrt{\frac{2240 \times 10}{7 \times 2}}=40 \mathrm{~m} / \mathrm{sec} \end{aligned} $
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be $\frac{x}{7}$. The value of $x$ is ___________.
Explanation:
Solid Sphere :
$ \begin{aligned} & I_{\text {tangent }}=I_{\mathrm{cm}}+m R^{2} \\\\ & =\frac{2}{5} m R^{2}+m R^{2}=\frac{7}{5} m R^{2} \\\\ & =7 R^{2} \quad(m=5 \mathrm{~kg}) \end{aligned} $
CIRCULAR DISC :
$ \begin{aligned} I_{\text {disc }} & =I_{\mathrm{cm}}+m R^{2} \\\\ & =\frac{m R^{2}}{4}+m R^{2} \\\\ & =\frac{5}{4} m R^{2} \\\\ & =5 R^{2} \end{aligned} $
$\frac{l_{\text {disc }}}{l_{\text {tangent }}}=\frac{5}{7}$
Concept :
1. For Solid Sphere :
Position of the axis of rotation : About its diametric axis which passes through its centre of mass
Moment of Inertia (I) = ${2 \over 5}M{R^2}$
Radius of gyration (K) = $\sqrt {{2 \over 5}} R$
Position of the axis of rotation : About a tangent to the sphere
Moment of Inertia (I) = ${7 \over 5}M{R^2}$
Radius of gyration (K) = $\sqrt {{7 \over 5}} R$
2. For CIRCULAR DISC :
Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre
Moment of Inertia (I) = ${1 \over 2}M{R^2}$
Radius of gyration (K) = ${R \over {\sqrt 2 }}$
Position of the axis of rotation : About the diametric axis
Moment of Inertia (I) = ${1 \over 4}M{R^2}$
Radius of gyration (K) = ${R \over { 2 }}$
$\mathrm{I_{CM}}$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $\mathrm{I_{AB}}$ is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance $\frac{2}{3}$R from center. Where R is the radius of the disc. The ratio of $\mathrm{I_{AB}}$ and $\mathrm{I_{CM}}$ is $x:9$. The value of $x$ is _____________.
Explanation:
$ =\frac{1}{2} M R^{2}+\frac{4}{9} M R^{2} $
$ \begin{aligned} & =\frac{(9+8) M R^{2}}{18}=\left(\frac{17}{18}\right) M R^{2} \end{aligned} $
$\frac{I_{A B}}{I_{\mathrm{cm}}}=\frac{17 / 18}{1 / 2}=\left(\frac{17}{9}\right)$
Value of $x=17$
A uniform solid cylinder with radius R and length L has moment of inertia I$_1$, about the axis of the cylinder. A concentric solid cylinder of radius $R'=\frac{R}{2}$ and length $L'=\frac{L}{2}$ is carved out of the original cylinder. If I$_2$ is the moment of inertia of the carved out portion of the cylinder then $\frac{I_1}{I_2}=$ __________.
(Both I$_1$ and I$_2$ are about the axis of the cylinder)
Explanation:
$ \begin{aligned} & I_{2}=\frac{\left[\rho \pi\left(\frac{R}{2}\right)^{2} \frac{L}{2}\right]\left(\frac{R}{2}\right)^{2}}{2} \\\\ & \therefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1} \end{aligned} $
Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be $\sqrt x$ cm. The value of $x$ is ________.
Explanation:
For radius of gyration $\mathrm{I}_{\mathrm{PQ}}=\mathrm{mk}^2$
$ \begin{aligned} & \mathrm{k}^2=\frac{2}{5} \mathrm{R}^2+(10 \mathrm{~cm})^2 \\\\ & =\frac{2}{5}(5)^2+100 \\\\ & =10+100=110 \\\\ & \mathrm{k}=\sqrt{110} \mathrm{~cm} \\\\ & \mathrm{x}=110 \end{aligned} $
Four identical discs each of mass '$\mathrm{M}$' and diameter '$\mathrm{a}$' are arranged in a small plane as shown in figure. If the moment of inertia of the system about $\mathrm{OO}^{\prime}$ is $\frac{x}{4} \,\mathrm{Ma}^{2}$. Then, the value of $x$ will be ____________.
Explanation:
$I = 2 \times \left( {{{M{{\left( {{a \over 2}} \right)}^2}} \over 4}} \right) + 2 \times \left( {{{M{{\left( {{a \over 2}} \right)}^2}} \over 4} + M{{\left( {{a \over 2}} \right)}^2}} \right)$
$ = {{M{a^2}} \over 8} + {{5M{a^2}} \over 8}$
$ = {{6M{a^2}} \over 8} = {3 \over 4}M{a^2}$
A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $4 \mathrm{~ms}^{-1}$, is ____________ cm. (take g = $10 \mathrm{~ms}^{-2}$)
Explanation:
For a solid cylinder, the moment of inertia I is given by $\frac{1}{2} m r^2$. The kinetic energy due to rotation is given by $\frac{1}{2} I \omega^2$. But we also know that $\omega = \frac{v}{r}$, hence the rotational kinetic energy can be written as $\frac{1}{2} \frac{1}{2} m v^2 = \frac{1}{4} m v^2$, where $\frac{1}{2}$ is from the moment of inertia of the solid cylinder.
Therefore, the total kinetic energy (linear + rotational) when the string snaps is $\frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2$.
Equating this to the potential energy $mgh$ and solving for h gives the result :
$ h = \frac{v^2}{2g} \times \frac{3}{2} = 1.2 \, \mathrm{m} = 120 \, \mathrm{cm} $
Alternate Method :
Applying COE, we get
$ m g h=\frac{1}{2} m v^2\left(1+\frac{\mathrm{K}^2}{r^2}\right) $
$\mathrm{K}=$ radius of gyration
For a solid cylinder, $\frac{\mathrm{K}^2}{r^2}=\frac{1}{2}$
$ \begin{aligned} \therefore h & =\frac{v^2}{2 g}\left(1+\frac{1}{2}\right) \\\\ & =\frac{16}{2 \times 10} \times \frac{3}{2} \\\\ & =1.2 \mathrm{~m} \\\\ & =120 \mathrm{~cm} \end{aligned} $
A pulley of radius $1.5 \mathrm{~m}$ is rotated about its axis by a force $F=\left(12 \mathrm{t}-3 \mathrm{t}^{2}\right) N$ applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $4.5 \mathrm{~kg} \mathrm{~m}^{2}$, the number of rotations made by the pulley before its direction of motion is reversed, will be $\frac{K}{\pi}$. The value of K is ___________.
Explanation:
$FR = I\alpha $
$\alpha = {{(12t - 3{t^2}) \times 1.5} \over {4.5}} = 4t - {t^2}$
$w = \int {\alpha \,dt = 2{t^2} - {{{t^3}} \over 3}} $
$w = 0$
$ \Rightarrow {t^2}\left[ {2 - {t \over 3}} \right] = 0$
$t = 6$ sec
$\left. {\theta = \int\limits_0^6 {\left[ {2{t^2} - {{{t^3}} \over 3}} \right]dt = \left[ {{{2{t^3}} \over 3} - {{{t^4}} \over {12}}} \right]} } \right|_0^6$
$ = \left[ {{2 \over 3} \times {6^3} - {{{6^4}} \over {12}}} \right] = 36$
$n = {{36} \over {2\pi }}$
$ = {{18} \over \pi }$
The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ___________ $\mathrm{m}$.
Given, the length of the rod is $10 \sqrt{3} \mathrm{~m}$.
Explanation:
$l = {{M{L^2}} \over {12}} = M{K^2}$
$K = {L \over {\sqrt {12} }} = {{10\sqrt 3 } \over {\sqrt {12} }} = 5\,m$
A disc of mass $1 \mathrm{~kg}$ and radius $\mathrm{R}$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $4 \sqrt{\frac{x}{3 R}} \,\operatorname{rad}{s}^{-1}$ where $x=$ ____________. $\left(g=10 \mathrm{~ms}^{-2}\right)$
Explanation:
Loss in P.E. = Gain in K.E.
$2mgR = {1 \over 2}\left[ {{1 \over 2}m{R^2} + m{R^2}} \right]{w^2}$
$2mgR = {1 \over 2} \times {3 \over 2}m{R^2}\,{w^2}$
${w^2} = {{8g} \over {3R}}$
$w = \sqrt {{{8g} \over {3R}}} = 4\sqrt {{g \over {2 \times 3R}}} $
$ \Rightarrow x = {g \over 2} = 5$
Four particles with a mass of 1 kg, 2 kg, 3 kg and 4 kg are situated at the corners of a square with side 1 m (as shown in the figure). The moment of inertia of the system, about an axis passing through the point O and perpendicular to the plane of the square, is ______________ kg m2.
Explanation:
$ \begin{aligned} &\mathrm{I}_{\text {net }}=(1+2+3+4) \cdot\left(\frac{a}{\sqrt{2}}\right)^{2} \text {, where a = side of square } \\\\ &=10 \times \frac{1^{2}}{2}=5 \mathrm{~kg} \mathrm{~m}^{2} \end{aligned} $
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If ${{{I_1}} \over {{I_2}}}$ is ${{x{\pi ^2}} \over 3}$, then the value of x will be ____________.
Explanation:
${I_1} = {{M{L^2}} \over 3}$ ..... (1)
For ring : ${I_2} = {{M{R^2}} \over 2}$
and $2\pi R = L$
$ \Rightarrow {I_2} = {M \over 2}\left( {{{{L^2}} \over {4{\pi ^2}}}} \right)$ ...... (2)
$ \Rightarrow {{{I_1}} \over {{I_2}}} = {{8{\pi ^2}} \over 3}$
$ \Rightarrow x = 8$
A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ____________ N.
(Take g = 10 ms$-$2)
Explanation:
20 $-$ T = 2a
and 0.1 $\times$ T = 0.02 $\alpha$ = ${{0.02a} \over {0.1}}$
T = 2a
$\Rightarrow$ a = 5 m/sec2
So T = 10 N
The position vector of 1 kg object is $\overrightarrow r = \left( {3\widehat i - \widehat j} \right)m$ and its velocity $\overrightarrow v = \left( {3\widehat j + \widehat k} \right)m{s^{ - 1}}$. The magnitude of its angular momentum is $\sqrt x $ Nm where x is ___________.
Explanation:
$\left| {\overrightarrow i } \right| = \left| {\overrightarrow r \times (m\overrightarrow v )} \right|$
$ = \left| {(3\widehat i - \widehat j) \times (3\widehat j + \widehat k)} \right|$
$ = \left| { - \widehat i - 3\widehat j + 9\widehat k} \right|$
$ = \sqrt {91} $
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be ${1 \over 2}\sqrt {xgh} $ m/s. The value of x is ___________.
Explanation:
For rolling wheel
$[12g\sin \alpha - 3g\sin \alpha ] \times R = (2 \times 12{R^2} + 3{R^2}) \times {a \over R}$
$ \Rightarrow {{9g\sin \alpha } \over {27}} = a$
$ \Rightarrow a = {{g\sin \alpha } \over 3}$
$\therefore$ $v = \sqrt {2 \times {{g\sin \alpha } \over 3} \times {h \over {\sin \alpha }}} = \sqrt {{2 \over 3}gh} $
$ = {1 \over 2} \times \sqrt {{8 \over 3}gh} $
$\therefore$ $x = {8 \over 3} = 2.67$
Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:
I1 = M.I. of solid sphere about its diameter
I2 = M.I. of solid cylinder about its axis
I3 = M.I. of solid circular disc about its diameter
I4 = M.I. of thin circular ring about its diameter
If 2(I2 + I3) + I4 = x . I1, then the value of x will be __________.
Explanation:
$2\left( {{1 \over 2} + {1 \over 4}} \right) \times M{(2R)^2} + {1 \over 2}M{(2R)^2} = x{2 \over 5}M{(2R)^2}$
$ \Rightarrow 1 + {1 \over 2} + {1 \over 2} = x \times {2 \over 5}$
$ \Rightarrow x = 5$
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x $\times$ 10$-$2 kg. The value of x is ___________.
Explanation:
Let $\mathrm{M}$ be the mass of the meter scale.
The weight $\mathrm{Mg}$ of the scale acts at $50 \mathrm{~cm}$ mark.
Finally after putting two coins on the meter scale, balancing the torques about the knife edge, we get,
$20 g \times 30=\mathrm{Mg} \times 10$
$ \begin{aligned} M & =60 \mathrm{~g} \\\\ & =60 \times 10^{-3} \mathrm{~kg} \\\\ & =6 \times 10^{-2} \mathrm{~kg} \\\\ & =x \times 10^{-2} \mathrm{~kg} \end{aligned} $
On comparing, we get $x=6$
Explanation:
by energy conservation $mgl = {1 \over 2}I{\omega ^2} = {1 \over 2}{{m{l^2}{\omega ^2}} \over 3}$
$ \Rightarrow \omega = \sqrt {{{6g} \over l}} $
As we know the relation between the linear speed and angular speed,
$v = \omega r = \omega l = \sqrt {6gl} $
$v = \sqrt {6 \times 10 \times .6} $ = 6 m/s
Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.
If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, ${r \over 2}$ distance from the end A of the handle will be ................ Mr2.
Explanation:
$I = \left[ {{I_1} + M{{\left( {{5 \over 2}r} \right)}^2}} \right] + \left[ {{I_2} + M{{\left( {{{13r} \over 2}} \right)}^2}} \right]$
$ = \left[ {{{M(36{r^2})} \over {12}} + {{M(25{r^2})} \over 4}} \right] + \left[ {{{M{r^2}} \over 2} + {{169M{r^2}} \over 4}} \right]$
= 52 Mr2
Ans. 52.00
Explanation:
${1 \over 2}{I_1}{({\omega _1})^2} = {1 \over 2}{I_2}{({\omega _2})^2}$
${I_1}{\left( {{v \over {3R}}} \right)^2} = {I_2}{\left( {{v \over R}} \right)^2}$
${{{I_1}} \over {{I_2}}} = {9 \over 1}$
Explanation:
$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$
$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$
$ = 0.4\pi = 4\pi \times {10^{ - 2}}$
$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$
Where $\alpha$ and $\beta$ are dimensional constants.
The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.
Explanation:
$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$
$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$
$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$
$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$
At t = 0, $\overrightarrow L = \overrightarrow 0 $
$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$
t $-$ 2 (t $-$ 5) = 0
t = 10 sec
Explanation:
$\left| \omega \right| = {{{v_0}} \over R}$
${\overrightarrow v _p} = {v_0}\widehat i + \omega R( - \widehat j) = {v_0}\widehat i - {v_0}\widehat j$
$\left| {{{\overrightarrow v }_p}} \right| = \sqrt 2 {v_0}$
$x = 02$
Explanation:
Ring
mgh = ${1 \over 2}I{\omega ^2}$
mgh = $ = {1 \over 2}(2m{R^2}){{v_R^2} \over {{R^2}}}$
${v_R} = \sqrt {gh} $
Solid cylinder
mgh = ${1 \over 2}I{\omega ^2}$
mgh $ = {1 \over 2}\left( {{3 \over 2}m{R^2}} \right){{v_C^2} \over {{R^2}}}$
${v_C} = \sqrt {{{4gh} \over 3}} $
${{{v_R}} \over {{v_C}}} = {{\sqrt 3 } \over 2}$
Explanation:
$\alpha = 1200 \times 6$
$\theta = {\omega _0}t + {1 \over 2}\alpha {t^2}$
$ = 600 \times {{10} \over {60}} + {1 \over 2} \times 1200 \times 6 \times {1 \over {36}}$
$\theta = 200$
Explanation:
When the disc slips down the inclined plane, it takes time t1. Therefore, in this case its acceleration, a1 = g sin$\theta$
$\because$ s = ut1 + ${1 \over 2}$a1t$_1^2$
$\Rightarrow$ s = ${1 \over 2}$ g sin$\theta$t$_1^2$ .... (i)
And when the disc rolls down the inclined plane, it takes time t2. Therefore in this case, its acceleration,
${a_2} = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}} = {{g\sin \theta } \over {1 + {1 \over 2}}} = {2 \over 3}g\sin \theta $ [$\because$ for disc, ${{{K^2}} \over {{R^2}}} = {1 \over 2}$]
$\because$ s = ut2 + ${1 \over 2}$ a2t$_2^2$ = ${1 \over 2}$ . ${2 \over 3}$ g sin$\theta$ t$_2^2$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
${{{t_2}} \over {{t_1}}} = \sqrt {{3 \over 2}} $ .... (iii)
According to question, value of ${{{t_2}} \over {{t_1}}}$ is $\sqrt {{3 \over x}} $.
Comparing it with Eq. (iii), we get x = 2
Explanation:
$ = 82\pi $
${\omega _i} = {{900 \times 2\pi } \over {60}} = 30\pi $
$\alpha = {{{\omega _f} - {\omega _i}} \over t}$
$ = {{82\pi - 30\pi } \over {26}}$
= 2 $\pi$ rad/sec2
$\theta = {{\omega _f^2 - \omega _i^2} \over {2\alpha }}$
$ = {{(82\pi + 30\pi )(82\pi - 30\pi )} \over {2 \times 2\pi }}$
$ = {{(112 \times 52){\pi ^2}} \over {4\pi }}$
No. of revolution $ = {{(112 \times 13)\pi } \over {2\pi }}$
= 728
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]
Explanation:
IR = mR2, aR = g sin$\theta$/2
ID = ${{m{R^2}} \over 2}$, aD = ${2 \over 3}$ g sin$\theta$
ISC = ${{m{R^2}} \over 2}$, aSC = ${2 \over 3}$ g sin$\theta$
ISC = ${2 \over 5}$mR2, aSS = ${5 \over 7}$ g sin$\theta$
S = ut + ${1 \over 2}$at2,
t = $\sqrt {{{2S} \over a}} $
$ \therefore $ t $\propto$ ${1 \over {\sqrt a }}$
solid sphere will take minimum time.
Explanation:
We know that, on an inclined plane
Acceleration, $a = {{g\sin \theta } \over {1 + {I \over {m{R^2}}}}}$
$ \Rightarrow a = {{g\sin \theta } \over {1 + {1 \over 2}}}$ [$\because$ For disc, $I = {{m{R^2}} \over 2}$]
$ \Rightarrow a = {2 \over 3}g\sin \theta $ ....... (i)
As per question, acceleration of the disc will be ${2 \over b}g\sin \theta $.
Comparing it with Eq. (i), we get
$b = 3$
Explanation:
$ = \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)$
$ = ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)$
$ = - 16\widehat j + 12\widehat k$
$|\overrightarrow \tau |\, = 20$ units
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s$-$1.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]
Explanation:
$\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10$ rad/s2
${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta $
${(50)^2} = {0^2} + 2(10)\Delta \theta $
$ \Rightarrow \Delta \theta = {{2500} \over {20}}$
$\Delta \theta = 125$ rad
No. of revolution $ = {{125} \over {2\pi }} \approx 20$ revolution
Suppose the force is $\overrightarrow P $ resolved parallel to the arms AB and AC of the frame.
The magnitude of the resolved component along the arm AC is xN.
The value of x, to the nearest integer, is ___________.
[Given : sin(35$^\circ$) = 0.573, cos(35$^\circ$) = 0.819
sin(110$^\circ$) = 0.939, cos(110$^\circ$) = $-$ 0.342 J
Explanation:
Component along AC
= 100 cos 35$^\circ$ N
= 100 $\times$ 0.819 N
= 81.9 N
$ \approx $ 82 N
Explanation:
MOI of AB about $P:{I_{ABp}} = {{{M \over 6}{{\left( {{l \over 6}} \right)}^2}} \over {12}}$
MOI of AB about O,
${I_{A{B_O}}} = \left[ {{{{M \over 6}{{\left( {{l \over 6}} \right)}^2}} \over {12}} + {M \over 6}{{\left( {{l \over 6}{{\sqrt 3 } \over 2}} \right)}^2}} \right]$
${I_{Hexago{n_0}}} = 6{I_{A{B_0}}} = M\left[ {{{{l^2}} \over {12 \times 36}} + {{{l^2}} \over {36}} \times {3 \over 4}} \right]$
$ = {6 \over {100}}\left[ {{{24 \times 24} \over {12 \times 36}} + {{24 \times 24} \over {36}} \times {3 \over 4}} \right]$
= 0.8 kgm2
= 8 $\times$ 10$-$1 kg-m2
Explanation:
The given situation is shown in the following figure,
Applying law of conservation of angular momentum about pivotal point,
${L_i} = {L_f} \Rightarrow mvl = I\omega $
$ \Rightarrow mvl = \left( {{1 \over 3}M{L^2} + m{L^2}} \right)\omega $ ..... (i)
Given, m = 0.1 kg, v = 80 ms$-$1, M = 0.9 kg, l = 1 m
Substituting these values in eq. (i), we get
$0.1 \times 80 \times 1 = \left( {{{0.9 \times {1^2}} \over 3} + 0.1 \times {1^2}} \right)\omega $
$8 = \left( {{3 \over {10}} + {1 \over {10}}} \right)\omega \Rightarrow 8 = {4 \over {10}}\omega \Rightarrow \omega = 20$ rad/s
$\left( {4\widehat i + 3\widehat j - \widehat k} \right)$ m. Then the magnitude of torque
about the point $\left( {\widehat i + 2\widehat j + \widehat k} \right)$ m will be $\sqrt x $ N m.
The value of x is _______.
Explanation:
$ = \left| {\matrix{ i & j & k \cr 3 & 1 & { - 2} \cr 1 & 2 & 3 \cr } } \right|$
$ = \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)$
$\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k$
$\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195} $
$ \therefore $ $x = 195$
Explanation:
${I_i} = {{M{R^2}} \over 2}$
${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$
$ = {5 \over 4}.{{M{R^2}} \over 2}$
$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \right).\omega $
$ \Rightarrow $ $\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} \right)} \right]\omega ' = {{M{R^2}} \over 2}\omega $
$\omega = {4 \over 5}\omega $
loss of K.E. = ${{Loss} \over {{K_i}}} \times 100 $
= ${{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} \right){{\left( {{4 \over 5}\omega } \right)}^2}} \over {{1 \over 2}I{\omega ^2}}}$ $ \times $ 100
= ${{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} \right)} \over {{\omega ^2}}}$ $ \times $ 100 = $\left( {1 - {{80} \over {100}}} \right) \times 100$
= 20%
Explanation:
I = m$a$2 + ${{m{a^2}} \over 4}$ = ${5 \over 4}m{a^2}$
Accoeding to the question,
${5 \over 4}m{a^2} = {N \over {20}}m{a^2}$
$ \Rightarrow $ N = 25
Explanation:
$ \Rightarrow $ $\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$
$ \Rightarrow $ $\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$
$ \Rightarrow $ $\left( {1 + {{2 \times 80} \over {200}}} \right){\omega _1} = {\omega _2}$
$ \Rightarrow $ ${\omega _2} = \left( {1.8} \right){\omega _1}$
$ \Rightarrow $ 2$\pi $f2 = 2$\pi $f1 $ \times $ 1.8
$ \Rightarrow $ f2 = 5 $ \times $ 1.8 = 9
(Breaking stress of wire = 4.8 × 107 Nm–2 and
area of cross-section of the wire = 10–2 cm2) is:
Explanation:
Breaking stress = $\sigma $ = ${{m{\omega ^2}l} \over A}$
$ \Rightarrow $ ${\omega ^2}$ = ${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$ = 16
$ \Rightarrow $ $\omega $ = 4