Rotational Motion
270 Questions
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Q226
Allen
JEE Main Practice Paper
MCQ
A solid sphere with a velocity (of centre of mass) v and angular velocity $\omega$ is gently placed on a rough horizontal surface. The frictional force on the sphere:
A.
must be forward (in direction of v)
B.
must be backward (opposite to v)
C.
cannot be zero
D.
none of the above
Q227
Allen
JEE Main Practice Paper
MCQ
A hoop and a solid cylinder have the same mass and radius. They both roll, without slipping, on a horizontal surface. If their kinetic energies are equal
A.
the hoop has a greater translational speed than the cylinder
B.
the cylinder has a greater translational speed than the hoop
C.
the hoop and the cylinder have the same translational speed
D.
the hoop has a greater rotational speed than the cylinder
Q228
Allen
JEE Main Practice Paper
MCQ
The moment of inertia of a solid cylinder about its axis is given by $(1/2)MR^{2}$ . If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A.
1 : 1
B.
2 : 2
C.
1 : 2
D.
1 : 3
Q229
Allen
JEE Main Practice Paper
MCQ
A rod of weight w is supported by two parallel knives at end points A and B and is in equilibrium in a horizontal position. The knives are at a distance 'd' from each other. The centre of mass of the rod is at a distance 'x' from A.
A.
the normal reaction at A is $\frac{wx}{d}$
B.
the normal reaction at A is $\frac{w(d-x)}{2d}$
C.
the normal reaction at B is $\frac{wx}{d}$
D.
the normal reaction at B is $\frac{w(d-x)}{d}$
Q230
Allen
JEE Main Practice Paper
MCQ
A man spinning in free space changes the shape of his body, eg. by spreading his arms or curling up. By doing this, he can not change his
A.
moment of inertia
B.
angular momentum
C.
angular velocity
D.
rotational kinetic energy
Q231
Allen
JEE Main Practice Paper
MCQ
A heavy seesaw (i.e., not massless) is out of balance. A light girl sits on the end that is tilted downward, and a heavy body sits on the other side so that the seesaw now balances. If they both move forward so that they are one-half their original distance from the pivot point (the fulcrum) what will happen to the seesaw?
A.
The side the body is sitting on will tilt downward
B.
The side the girl is sitting on will once again tilt downward
C.
Nothing; the seesaw will still be balanced
D.
It is impossible to say without knowing the masses and the distances
Q232
Allen
JEE Main Practice Paper
MCQ
A rod is hinged at its centre and rotated by applying a constant torque starting from rest. The power developed by the external torque as a function of time is:
A.
B.
C.
D.
Q233
Allen
JEE Main Practice Paper
MCQ
A hollow smooth uniform sphere A of mass āmā rolls without sliding on a smooth horizontal surface. It collides head on elastically with another stationary smooth solid sphere B of the same mass m and same radius. The ratio of kinetic energy of āBā to that of āAā just after the collision is :
A.
1:1
B.
2:3
C.
3:2
D.
None
Q234
Allen
JEE Main Practice Paper
MCQ
A wheel of radius r rolling on a straight line, the velocity of its centre being v. At a certain instant the point of contact of the wheel with the grounds is M and N is the highest point on the wheel (diametrically opposite to M). The incorrect statement is:
A.
The velocity of any point P of the wheel is proportional to MP.
B.
Points of the wheel moving with velocity greater than v form a larger area of the wheel than points moving with velocity less than v.
C.
The point of contact $M$ is instantaneously at rest.
D.
The velocities of any two parts of the wheel which are equidistant from centre are equal.
Q235
Allen
JEE Main Practice Paper
MCQ
Particles of masses 1, 2 and 3kg respectively are arranged at the corners of an equilateral triangle as shown in figure. Obtain the moment of inertia of the system about an axis passing through the centre of mass of the system and perpendicular to its plane. The side of the triangle is 1m.
A.
$4kg - m^{2}$
B.
$\frac{5}{6} kg\cdot m^{2}$
C.
$\frac{11}{6} kg - m^2$
C.
5 kg-mĀ²
Q236
Allen
JEE Main Practice Paper
MCQ
Inner and outer radii of a spool are r and R respectively. A thread is wound over its inner surface and placed over a rough horizontal surface. Thread is pulled by a force F as shown in fig. then in case of pure rolling:
A.
Thread unwinds, spool rotates anticlockwise and friction act leftwards.
B.
Thread winds, spool rotates clockwise and friction acts leftwards.
C.
Thread winds, spool moves to the right and friction act rightwards.
D.
Thread winds, spool moves to the right and friction does not come into existence.
Q237
Allen
JEE Main Practice Paper
MCQ
A yo-yo is resting on a perfectly rough horizontal table. Forces $F_{1}, F_{2}$ and $F_{3}$ are applied separately as shown. The correct statement is :-
A.
when $F_{3}$ is applied the centre of mass will move to the right.
B.
when $F_{2}$ is applied the centre of mass will move to the left.
C.
when $F_{1}$ is applied the centre of mass will move to the right.
D.
when $F_{2}$ is applied the centre of mass will move to the right.
Q238
Allen
JEE Main Practice Paper
MCQ
Pulley is rotating with angular velocity $\omega$ . Pulley is sufficiently rough so that string does not slip on the pulley. Then angular momentum of system (pulley + blocks) about axle of pulley is :-
A.
$I\omega$
B.
$(m_{1} + m_{2} + I)\omega r$
C.
$(m_{1} + m_{2})r^{2}\omega +I\omega$
D.
$(m_{1}r^{2} + m_{2}r^{2} + I)\omega^{2}$
Q239
Allen
JEE Main Practice Paper
MCQ
The free end of the string wound on the surface of a solid cylinder of mass M and radius r is pulled up by a force F as shown. If there is sufficient friction between cylinder and floor so that the cylinder is able to roll without slipping, find the maximum angular acceleration that the cylinder can have (without cylinder leaving contact with surface).
A.
$\frac{g}{r}$
B.
$\frac{2g}{3r}$
C.
$\frac{4g}{3r}$
Q240
Allen
JEE Main Practice Paper
MCQ
From a solid sphere of mass M and radius R, a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:-
A.
$\frac{4MR^{2}}{9\sqrt{3}\pi}$
B.
$\frac{4MR^{2}}{3\sqrt{3}\pi}$
C.
$\frac{MR^{2}}{32\sqrt{2}\pi}$
D.
$\frac{MR^{2}}{16\sqrt{2}\pi}$
Q241
Allen
JEE Main Practice Paper
MCQ
A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to :-
A.
turn left and right alternately
B.
turn left
C.
turn right
D.
go straight
Q242
Allen
JEE Main Practice Paper
Numerical
A rigid horizontal smooth rod AB of mass 0.75 kg and length 40 cm can rotate freely about a fixed vertical axis through its mid-point O. Two rings each of mass 1 kg are initially at rest at a distance of 10 cm from O on either side of the rod. The rod is set in rotation with an angular velocity of 30 radians per second. Find the velocity of each ring along the length of the rod in m/s when they reach the ends of the rod.
Correct Answer: 3
Explanation:
Ans. (3)
$\begin{array}{l} \left(\frac {M _ {r} \ell^ {2}}{1 2} + 2 m _ {P} (0. 1) ^ {2}\right) \omega_ {1} = \left(\frac {M _ {r} \ell^ {2}}{1 2} + 2 m _ {P} (0. 2) ^ {2}\right) \omega_ {2} \\ \Rightarrow \quad \left(\frac {0 . 7 5 \times 0 . 4 \times 0 . 4}{1 2} + 2 \times 1 \times (0. 1) ^ {2}\right) 3 0 = \left(\frac {0 . 7 5 \times 0 . 4 \times 0 . 4}{1 2} + 2 \times 1 \times (0. 2) ^ {2}\right) \omega_ {2} \\ \Rightarrow \quad \omega_ {2} = 1 0 r a d / s \\ K E _ {i} = K E _ {f} \\ \Rightarrow \quad \frac {1}{2} I _ {1} \omega_ {1} ^ {2} = \frac {I}{2} \omega_ {2} ^ {2} + 2 \times \frac {1}{2} M V ^ {2} \\ v = 3 m / s \end{array}$
$\begin{array}{l} \left(\frac {M _ {r} \ell^ {2}}{1 2} + 2 m _ {P} (0. 1) ^ {2}\right) \omega_ {1} = \left(\frac {M _ {r} \ell^ {2}}{1 2} + 2 m _ {P} (0. 2) ^ {2}\right) \omega_ {2} \\ \Rightarrow \quad \left(\frac {0 . 7 5 \times 0 . 4 \times 0 . 4}{1 2} + 2 \times 1 \times (0. 1) ^ {2}\right) 3 0 = \left(\frac {0 . 7 5 \times 0 . 4 \times 0 . 4}{1 2} + 2 \times 1 \times (0. 2) ^ {2}\right) \omega_ {2} \\ \Rightarrow \quad \omega_ {2} = 1 0 r a d / s \\ K E _ {i} = K E _ {f} \\ \Rightarrow \quad \frac {1}{2} I _ {1} \omega_ {1} ^ {2} = \frac {I}{2} \omega_ {2} ^ {2} + 2 \times \frac {1}{2} M V ^ {2} \\ v = 3 m / s \end{array}$
Q243
Allen
JEE Main Practice Paper
Numerical
A carpet of mass 'M' made of inextensible material is rolled along its length in the form of a cylinder of radius 'R' and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. The horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R/2 is $\left(\sqrt{\frac{ngR}{3}}\right)$ . Find n.
Correct Answer: 14
Explanation:
Ans. (14)
$w _ {f r} = 0$
$w _ {g} = K E _ {f}$
$m g R - \frac {m}{4} \frac {g R}{2} = \frac {1}{2} \frac {m}{4} v ^ {2} + \frac {1}{2} \frac {\frac {m}{4} \left(\frac {R}{2}\right) ^ {2}}{2} w ^ {2}$
$\frac {7}{8} m g R = m v ^ {2} \left(\frac {1}{8} + \frac {1}{1 6}\right) = m v ^ {2} \frac {3}{1 6}$
$\therefore \quad v = \sqrt {\frac {1 4}{3} g R}$
$w _ {f r} = 0$
$w _ {g} = K E _ {f}$
$m g R - \frac {m}{4} \frac {g R}{2} = \frac {1}{2} \frac {m}{4} v ^ {2} + \frac {1}{2} \frac {\frac {m}{4} \left(\frac {R}{2}\right) ^ {2}}{2} w ^ {2}$
$\frac {7}{8} m g R = m v ^ {2} \left(\frac {1}{8} + \frac {1}{1 6}\right) = m v ^ {2} \frac {3}{1 6}$
$\therefore \quad v = \sqrt {\frac {1 4}{3} g R}$
Q244
Allen
Rotational Motion
Numerical
A rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then find the maximum angle 'Īø'(in degree) made by the hinged upper half with the vertical.
Correct Answer: 60
Explanation:
Ans. (60)
$\begin{array}{l} m g \frac {\ell}{2} = \frac {1}{2} \frac {m \ell^ {2}}{3} \omega^ {2} \\ \therefore \quad \omega^ {2} = \frac {3 g}{\ell} \text {at bottom} \\ W _ {g} = \Delta K E \\ \Rightarrow \quad - \frac {m}{2} g \frac {\ell}{4} (1 - \cos \theta) = 0 - \frac {1}{2} \left(\frac {m}{2}\right) \frac {\ell^ {2}}{4} \times \frac {1}{3} \omega^ {2} \\ \therefore \quad \frac {m g \ell}{8} (1 - \cos \theta) = \frac {m g \ell}{1 6} \\ \therefore \quad \cos \theta = \frac {1}{2} \end{array}$
$m a \cos \theta \times \frac {\ell}{2} = I \alpha = \left(\frac {1}{3} m \ell^ {2}\right) \alpha$
$\frac {\omega d \omega}{d \theta} = \frac {3 a \cos \theta}{2 \ell} \Rightarrow \int_ {0} ^ {\omega} \omega d \omega = \frac {3 a}{2 \ell} [ \sin \theta ] _ {0} ^ {\frac {\pi}{2}}$
$\omega = \sqrt {\frac {3 a}{\ell}}$
$v = \omega \ell = \ell \sqrt {\frac {3 a}{\ell}} = \sqrt {3 a \ell}$
$V = \sqrt {1 5} f t / s$
$\Rightarrow \alpha = \frac {\mu g}{\ell}$
$\omega_ {0} = \left(\frac {\mu g}{\ell}\right) t$
$\begin{array}{l} m g \frac {\ell}{2} = \frac {1}{2} \frac {m \ell^ {2}}{3} \omega^ {2} \\ \therefore \quad \omega^ {2} = \frac {3 g}{\ell} \text {at bottom} \\ W _ {g} = \Delta K E \\ \Rightarrow \quad - \frac {m}{2} g \frac {\ell}{4} (1 - \cos \theta) = 0 - \frac {1}{2} \left(\frac {m}{2}\right) \frac {\ell^ {2}}{4} \times \frac {1}{3} \omega^ {2} \\ \therefore \quad \frac {m g \ell}{8} (1 - \cos \theta) = \frac {m g \ell}{1 6} \\ \therefore \quad \cos \theta = \frac {1}{2} \end{array}$
$m a \cos \theta \times \frac {\ell}{2} = I \alpha = \left(\frac {1}{3} m \ell^ {2}\right) \alpha$
$\frac {\omega d \omega}{d \theta} = \frac {3 a \cos \theta}{2 \ell} \Rightarrow \int_ {0} ^ {\omega} \omega d \omega = \frac {3 a}{2 \ell} [ \sin \theta ] _ {0} ^ {\frac {\pi}{2}}$
$\omega = \sqrt {\frac {3 a}{\ell}}$
$v = \omega \ell = \ell \sqrt {\frac {3 a}{\ell}} = \sqrt {3 a \ell}$
$V = \sqrt {1 5} f t / s$
$\Rightarrow \alpha = \frac {\mu g}{\ell}$
$\omega_ {0} = \left(\frac {\mu g}{\ell}\right) t$
Q245
Allen
Rotational Motion
Numerical
The door of an automobile is open and perpendicular to the body. The automobile starts with an acceleration of $2 \, ft/sec^{2}$ , and the width of the door is 30 inches. Treat the door as a uniform rectangle, and neglect friction. The speed of its outside edge as seen by the driver when the door closes is $\sqrt{n} \, ft/sec$ . Find n.
Correct Answer: 15
Explanation:
15
Q246
Allen
Rotational Motion
Numerical
A dumbbell consists of two identical particles of mass m connected by a rigid light rod of length $2\ell$ . The dumbbell is set spinning with angular speed $\omega_{0}$ on a surface with a small friction coefficient $\mu_{k}$ . If dumbbell stops in time $t=\frac{K\omega_{0}\ell}{2\mu g}$ where K is a constant, then find the value of K.
Correct Answer: 2
Explanation:
2
Q247
Allen
Rotational Motion
Numerical
A uniform circular disc has radius R and mass m. A particle also of mass m is fixed at a point A on the edge of the disc as in figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R/4 from the centre C of the disc. The line AC is perpendicular to PQ. Initially the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find the linear speed (in m/s) of the particle at it reaches its lowest position. Take radius R = 50 cm.
Correct Answer: 5
Explanation:
Ans. (5)
$K E _ {f} = \frac {I}{2} I \omega^ {2}; K E _ {i} = 0; W _ {g} = m g \left(\frac {2 R}{4}\right) + m g 2 \times \left(\frac {5 R}{4}\right)$
$I = m \left(\frac {5 R}{4}\right) ^ {2} + \left(\frac {M R ^ {2}}{4} + \frac {M R ^ {2}}{1 6}\right) = M R ^ {2} \left(\frac {2 5}{1 6} + \frac {1}{4} + \frac {1}{1 6}\right)$
$W _ {g} = K E _ {f} - K E _ {i}$
$\text { Get } v = \omega \frac {5 R}{4} = 5 m / s$
$K E _ {f} = \frac {I}{2} I \omega^ {2}; K E _ {i} = 0; W _ {g} = m g \left(\frac {2 R}{4}\right) + m g 2 \times \left(\frac {5 R}{4}\right)$
$I = m \left(\frac {5 R}{4}\right) ^ {2} + \left(\frac {M R ^ {2}}{4} + \frac {M R ^ {2}}{1 6}\right) = M R ^ {2} \left(\frac {2 5}{1 6} + \frac {1}{4} + \frac {1}{1 6}\right)$
$W _ {g} = K E _ {f} - K E _ {i}$
$\text { Get } v = \omega \frac {5 R}{4} = 5 m / s$
Q248
Allen
Rotational Motion
Numerical
Two men, each of mass 75 kg, stand on the rim of a horizontal large disc, diametrically opposite to each other. The disc has a mass 450 kg and is free to rotate about its axis. Each man simultaneously start along the rim clockwise with the same speed and reaches their original starting points on the disc. The angle turned by the disc with respect to the ground in this duration is $\frac{n\pi}{5}$ . Find n.
Correct Answer: 4
Explanation:
Ans. (4)
$2 m v R = \frac {M _ {1} R ^ {2}}{2} \omega_ {\mathrm{d}} \Rightarrow \frac {v}{R} = \frac {3}{2} \omega_ {d}$
$\therefore \quad \frac {\omega_ {m}}{\omega_ {D}} = \frac {3}{2}$
$\omega_ {m / D} = \frac {5 \omega_ {D}}{2}$
$\omega_ {m / D} t = 2 \pi$
$\omega_ {D} t = 4 \pi / 5$
$2 m v R = \frac {M _ {1} R ^ {2}}{2} \omega_ {\mathrm{d}} \Rightarrow \frac {v}{R} = \frac {3}{2} \omega_ {d}$
$\therefore \quad \frac {\omega_ {m}}{\omega_ {D}} = \frac {3}{2}$
$\omega_ {m / D} = \frac {5 \omega_ {D}}{2}$
$\omega_ {m / D} t = 2 \pi$
$\omega_ {D} t = 4 \pi / 5$
Q249
Allen
Rotational Motion
Numerical
A spool of inner radius R and outer radius 3R has a moment of inertia = $MR^{2}$ about an axis passing through its geometric centre, where M is the mass of the spool. A thread wound on the inner surface of the spool is pulled horizontally with a constant force = Mg. Find the acceleration (in $m/s^{2}$ ) of the point on the thread which is being pulled assuming that the spool rolls purely on the floor.
Correct Answer: 16
Explanation:
Ans. (16)
T - f = m a
$T R + f 3 R = m R ^ {2} \alpha ; a = \alpha 3 R$
Answer is $a + \alpha R$
T - f = m a
$T R + f 3 R = m R ^ {2} \alpha ; a = \alpha 3 R$
Answer is $a + \alpha R$
Q250
Allen
Rotational Motion
Numerical
A hollow cylinder with inner radius R, outer radius 2R and mass M is rolling with speed of its axis v. Its kinetic energy is $\frac{n}{16}Mv^{2}$ . Find n
Correct Answer: 13
Explanation:
Ans. (13)
$KE=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2};$ get I by subtraction. ( $I_{outer}-I_{inner}$ ).
$\omega = \frac {v}{2 R}$
$KE=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2};$ get I by subtraction. ( $I_{outer}-I_{inner}$ ).
$\omega = \frac {v}{2 R}$