Tag: rotational motion of a rigid body and moment of inertia

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Questions Related to rotational motion of a rigid body and moment of inertia

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 solid cylinder (SC),Hollow cylinder (HC)& solid sphere (SS)of same mass & radii are released simultaneously from the same height on an incline. The order in which they will reach the bottom is (From least time to most time order)

  1. SC,HC,SS

  2. SS,SS,HC

  3. SS,SC,HC

  4. HC,SC,SS
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Acceleration a = g*sin(theta) / (1 + I/MR^2). The object with the smallest I/MR^2 has the largest acceleration and reaches the bottom first. I/MR^2 values: SS (2/5), SC (1/2), HC (1). Order: SS, SC, HC.

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 solid homogeneous cylinder of height h and base radius r is kept vertically on a conveyer belt moving horizontally with an increasing velocity $v=a+{ bt }^{ 2 }$. If the cylinder is not allowed to slip then the time when the cylinder is about to topple, will be equal to

  1. $\dfrac { 2rg }{ bh } $

  2. $\dfrac { rg }{ bh } $

  3. $\dfrac { 2bg }{ rh } $

  4. $\dfrac { rg }{ 2bh } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Toppling occurs when the torque due to pseudo-force (ma) about the edge exceeds the torque due to gravity. The condition is m*a*h/2 = m*g*r. With a = d^2x/dt^2 = 2bt, we get m*(2bt)*h/2 = m*g*r, so t = rg/bh.

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 radius R rolls on a horizontal surface with linear velocity $ \overrightarrow {v} = v \hat {i} $ and angular velocity $ \overrightarrow {\omega} = - \omega \hat k $ there is a particle P on the circumference of the disc which has velocity in vertical direction. the height of that particle from the ground will be

  1. $ R + \dfrac {v}{ \omega} $

  2. $ R - \dfrac {v}{ \omega} $

  3. $ R + \dfrac {R}{ 2} $

  4. $ R - \dfrac {R}{ 2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The velocity of a point on the rim is v_p = v_cm + omega x r. For vertical velocity, the horizontal components must cancel. The height y = R - v/omega is a standard result for the point where horizontal velocity is zero.

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.