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Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A wheel of mass M and radius a and M.I. $I _G$ (about centre of mass) is set rolling with angular velocity $\omega$ up a rough inclined plane of inclination $\theta$. The distance travelled by it up the plane is :

  1. $\dfrac{I _G \omega^2}{2Mgsin\theta}$

  2. $\dfrac{\omega^2(Ma^2 + I _G}{2Mgsin\theta}$

  3. $\dfrac{I _G \omega}{Mgsin\theta}$

  4. $\dfrac{I _G \omega}{2Mgsin\theta}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using work-energy theorem: Initial KE = (1/2)Mv^2 + (1/2)I(omega)^2. Since v = omega*a, KE = (1/2)(M + I/a^2)v^2. The distance d = v^2 / (2*g*sin(theta)). Substituting v = omega*a gives the result.

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A disc of mass $m$ of radius $r$ is placed on a rough horizontal surface. A cue of mass $m$ hits the disc at a height $h$ from the axis passing through centre and parallel to the surface. The cue stop and falls down after impact. The disc starts pure rolling for

  1. $h < \dfrac{r}{3}$

  2. $h = \dfrac{r}{2}$

  3. $h > \dfrac{r}{2}$

  4. $h\ge \dfrac{r}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A uniform rigid rod has length $L$ and mass $m$. It lies on a horizontal smooth surface, and is rotated at a uniform angular velocity $\omega$ about a vertical axle passing through one of its ends. The force exerted by the axle on the rod will be

  1. $m \omega^2 L$ outward

  2. $m \omega^2 L$ inward

  3. $\dfrac{1}{2} m\omega^2 L$ outward

  4. $\dfrac{1}{2} m\omega^2 L$ inward
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

  1. Both reach at the same time

  2. Depends on their masses

  3. Disc

  4. Sphere
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Acceleration a = g*sin(theta) / (1 + I/MR^2). The sphere has I = (2/5)MR^2, so a = g*sin(theta) / 1.4. The disc has I = (1/2)MR^2, so a = g*sin(theta) / 1.5. The sphere has higher acceleration and arrives first.

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A body is given translational velocity and kept on a surface that has sufficient friction. Then:

  1. Body will move forward before pure rolling

  2. Body will move backward before pure rolling

  3. Body will start pure rolling immediately

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

since the body is given initial translational velocity so it will move forward while coming in pure rolling condition.

so the answer is A.

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

(a) A child stands at the center of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40$ rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

  1. $300\ rev/min\ and 2.5 E _1$

  2. <span>$100\ rev/min\ and 2.5 E _1$</span>

  3. <span>$500\ rev/min\ and 7.5 E _1$</span>

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By conservation of angular momentum, I1*omega1 = I2*omega2. Given I2 = (2/5)I1, then omega2 = (5/2)omega1 = 2.5 * 40 = 100 rev/min. KE2 = (1/2)I2*omega2^2 = (1/2)(2/5)I1(2.5*omega1)^2 = 2.5 * KE1.

Multiple choice physics force the turning of couple couple turning effect of force

Two small kids weighing 10 kg and 15 kg are trying to balance a seesaw of total length 5m, with the fulcrum at the centre. If one of the kids is sitting at an end, where should the other sit?

  1. $2.5 m$

  2. $1 m$

  3. $1.7 m$

  4. $2 m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a seesaw to balance, the torque on both sides must be equal. Let the fulcrum be at 0. One kid (10 kg) is at 2.5 m from the center. The other kid (15 kg) must be at distance x such that 10 * 2.5 = 15 * x. Solving for x gives 25 / 15 = 1.666... m, which is approximately 1.7 m.

Multiple choice physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

 If principle of moments for any object holds, then object is in state of

  1. inertia

  2. equilibrium

  3. suspension

  4. motion
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If principle of moments hold good, then the net torque about a given point is zero (usually CM or the pivoted point is zero). Hence the object does not rotate and is said to be in equilibrium

Multiple choice physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

A uniform dice of mass $10kg$ radius $1m$ is placed on a rought horizontal surface. The coefficient of friction between the disc and the surface is $0.2$. A horizontal time varying force is applied on the centre of the disc whose variation with time is shown in graph.
List-I                                                         List-IIDisc rolls without slipping                   at $t=7s$Disc rolls with slipping                       at $t=3s$  Disc starts slipping at                         at $t=4s$Friction force is $10N$ at              None

  1. $A-p,q;B-p;C-r;Dq$

  2. $A-p,r;B-s;C-s,p;D-q$

  3. $A-q,r;B-p;C-s;D-q$

  4. $A-p,q,r;B-q;r;C-s;p;D-p,q,r,s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

When slightly different weights are placed on the two pans of a beam balance, the beam comes to rest at an angle with the horizontal. The beam is supported at a single point P by a pivot. Then which of the following statement(s) is/are true ?

  1. The net torque about P due to the two weights is nonzero at the equilibrium position.

  2. The whole system does not continue to rotate about P because it has a large moment of inertia.

  3. The centre of mass of the system lies below P.

  4. The centre of mass of the system lies above P.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The whole system does not continue to rotate about P because the moment is balanced. Thus option B is wrong. And the center of mass of the system lies at pivot point P. Thus option C and D are wrong. As the force applied at the two points of suspension is different $\tau$ is different.