Rotational Motion
A spherical shell of 1 kg mass and radius R is rolling with angular speed $\omega$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is ${a \over 3}$ R2$\omega$. The value of a will be :
A ball is spun with angular acceleration $\alpha$ = 6t2 $-$ 2t where t is in second and $\alpha$ is in rads$-$2. At t = 0, the ball has angular velocity of 10 rads$-$1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :
A $\sqrt {34} $ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ef and Fw are the reaction forces of the floor and the wall, then ratio of ${F_w}/{F_f}$ will be :
(Use g = 10 m/s2.)
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Moment of inertia of solid sphere of radius R about any tangent. | (I) | ${5 \over 3}M{R^2}$ |
| (B) | Moment of inertia of hollow sphere of radius (R) about any tangent. | (II) | ${7 \over 5}M{R^2}$ |
| (C) | Moment of inertia of circular ring of radius (R) about its diameter. | (III) | ${1 \over 4}M{R^2}$ |
| (D) | Moment of inertia of circular disc of radius (R) about any diameter. | (IV) | ${1 \over 2}M{R^2}$ |
Choose the correct answer from the options given below :
One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through fixed end, then the elongation of the spring will be :
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is
A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads$-$1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads$-$1).
If force $\overrightarrow F = 3\widehat i + 4\widehat j - 2\widehat k$ acts on a particle position vector $2\widehat i + \widehat j + 2\widehat k$ then, the torque about the origin will be :
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 $x=6$
| List-I | List-II |
|---|---|
| (a) MI of the rod (length L, Mass M, about an axis $ \bot $ to the rod passing through the midpoint) | (i) $8M{L^2}/3$ |
| (b) MI of the rod (length L, Mass 2M, about an axis $ \bot $ to the rod passing through one of its end) | (ii) $M{L^2}/3$ |
| (c) MI of the rod (length 2L, Mass M, about an axis $ \bot $ to the rod passing through its midpoint) | (iii) $M{L^2}/12$ |
| (d) MI of the rod (Length 2L, Mass 2M, about an axis $ \bot $ to the rod passing through one of its end) | (iv) $2M{L^2}/3$ |
Choose the correct answer from the options given below:
Assertion A : Moment of inertia of a circular disc of mass 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be Ix, Iy and Iz respectively. The respectively radii of gyration about all the three axes will be the same.
Reason R : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below :
The correct statement for this situation is
[The coefficient of static friction, $\mu$s' is 0.4]
I1 = M.I. of thin circular ring about its diameter,
I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I3 = M.I. of solid cylinder about its axis and
I4 = M.I. of solid sphere about its diameter.
Then :
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