Physics

Rotational and Circular Motion

220 Questions

Rotational and circular motion examines the dynamics of objects moving in circular paths or rotating around an axis. Key concepts include angular momentum, torque, moment of inertia, and centripetal force. This is a highly scoring topic in the physics section of competitive exams.

Angular momentumCentripetal forceMoment of inertiaRolling objectsGyroscopic effect

Rotational and Circular Motion Questions

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A body is allowed to slide on a frictionless track from rest under-gravity.The track ends in a circular of diameter D. What should be the mini-mum height of the body in terms of D.          So that it may successfully complete the loop?

  1. $\dfrac{9}{4}D $
  2. $ \frac {5}{4} D $

  3. D

  4. 2 D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Velocity at top of the circle is $\sqrt{5gr}$ 

Energy lost to reach height D from height H  $ = mg(H-D)$

Energy gained in the form of $K.E= \dfrac12 mv^2= \dfrac12 m(5gr) = \dfrac{5mgD}{4}$ 

Equating K.E. and P.E. ,

$\dfrac{5mgD}{4} = mg(H-D)$

$H=\dfrac{9D}{4}$

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A body is a allowed to slide down a frictionless track from rest position at its top under gravity. The track ends in a circular loop of diameter $D$. Then, the minimum height of the inclined track  (in terms of $D$ ) so that it may complete successfully the loop is:

  1. $\dfrac{7D}{4}$
  2. $\dfrac{9D}{4}$
  3. $\dfrac{5D}{4}$
  4. $\dfrac{3D}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$V _{0}$ required for a complete vertical circle: $V _{0} =\sqrt{\dfrac{5}{2}gD}$
WET between top and bottom:
              $mgH  = \dfrac{1}{2}m V _{0}^{2}$


                 $gH  = \dfrac{5}{4}gD$

                   $H  = \dfrac{5}{4}D$

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A weightless thread can bear tension up to $3.7kg-wt$. A stone of mass $500g$ is tied to it and revolved in a circular path of radius $4$m in a verticle plane. If $g=10m/s^2$, then the maximum angular velocity of the stone will be:

  1. $2 rad/s$
  2. $4 rad/s$
  3. $16 rad/s$
  4. $\sqrt{21} rad/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Tension is maximum at lowest point of a verticle circle $T _{max}=mr\omega^2+mg$
$3.7g=0.5\times 4\times \omega^2+0.5g$
$2\omega^2=3.2g$
$\omega=\sqrt{1.6g}=\sqrt{1.6\times 10}$
$\omega=4rad/s$

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A body attached to a string of length describes a vertical circle such that it is just able to cross the highest point. Find the minimum velocity at the bottom of the circle.

  1. $\sqrt { 2gL } $
  2. $\sqrt { gL } $
  3. $\sqrt { 3gL } $
  4. $\sqrt { 5gL } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Radius of the circle traced will be equal to the length of the string
In circular motion tension in the string is balanced by the centripetal force
hence we know that,
$T=\dfrac { m{ v }^{ 2 } }{ r } $ 
$T=\dfrac { m{ v }^{ 2 } }{ L } $
Velocity at any point,$V=\sqrt { \dfrac { TL }{ m }  } $
And at lowest point the tension is balanced by the weight of the block
∴∴ tension = weight = mg
$V=\sqrt { \dfrac { mgL }{ m }  } =\sqrt { gL } $

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A small ball bearing is releases at the top of a long vertical column of glycerine of height $2h$. The ball bearing falls through a height $h$ in a time $t _{1}$ and then the remaining height with the terminal velocity in time $t _{2}$ Let $W _{1}$ and $W _{2}$ be the work done against viscous drag over these height. therefore.

  1. $t _{1}< t _{2}$
  2. $t _{1}> t _{2}$
  3. $W _{1}=W _{2}$
  4. $W _{1}< W _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics turning on a pivot the turning effect of a force moment of force or torque turning effect of force couple

A uniform meter rule is pivoted at its mid-point. A weight of $50gf$ suspended at one end of it. Where should a weight of $100 gf$ be suspended to keep the rule horizontal : 

  1. At distance $50 \ cm$ from the pivoted end.
  2. At distance $12.5 \ cm$ from the other end.
  3. At distance $37.5 \ cm$ from the other end.
  4. At distance $25 \ cm$ from the pivoted end.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As we have one meter rule pivoted at midpoint there will be a perpendicular distance of 50 cm from the contact of force of 50gf.
 Here we have moment of force of $r \times F$.
After substituting we get $moment\ of\ force = 50\times50 gf cm$.
Here this must be balanced by 100gf's moment of force.
 Let us think that there would be x perpendicular distance.
On equating we get 
$ 50\times50  gf cm = 100\times x gf cm$
$ x = 50\times 50/100 = 25 cm$. 

Hence $100 gf$ must be applied at a distance of $25 cm $.

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

A uniform metre rule of mass $100g$ is balanced on a fulcrum at mark $40cm$ by suspending an unknown mass $m$ at the mark $20cm$ . To which side the rule will tilt if the mass $m$ is moved to the mark $10cm$?

  1. on the side of mass $m$
  2. on the other side of mass $m$

  3. can't say

  4. it won't tilt
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As 100 g is balanced at fulcrum it will have a perpendicular distance of 40 cm from itself and mass m will have $100 - 20 = 80 cm $.


Here moment of forces are equal that is balance each other.Hence $ m\times 80 = 100\times  40 \Rightarrow $

$ m = 100/2 = 50 g $.

If there is mark at 10 cm then perpendicular distance would be 90 cm and moment of force would be $50\times90 = 4500\times g \ dyne - cm$ 
where as first moment of force is $100\times40 = 400\times g\ dyne- cm $. 

Hence it tilts on mass m side as it has more moment of force.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

A uniform rod AB of mass $3m$ and length $2l$ is lying at rest on a smooth horizontal table with a smooth vertical axis through the end $A$ . A particle of mass $2m$ moves with speed $2u$ across the table and strikes the rod at its mid point $C$. If the impact is perfectly elastic , then find the speed of the particle after impact if it strikes the rod normally 

  1. $\dfrac{7u}{3}$
  2. $\dfrac{2u}{3}$
  3. $\dfrac{u}{3}$
  4. $\dfrac{4u}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of angular momentum about the fixed axis A and the coefficient of restitution equation for the impact, the final velocities can be solved. For a rod of mass 3m and length 2l, the moment of inertia about A is (1/3)(3m)(2l)^2 = 4ml^2. Solving the system yields the particle's final speed as u/3.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

A disc of mass $100g$ and radius $10cm$ has a projection on its circumference. The mass of projection is negligible. A $20g$ bit of putty moving tangential to the disc with a velocity of $5m{s}^{-1}$ strikes the projection and sticks to it. The angular velocity of disc is

  1. $14.29rad{s}^{-1}$
  2. $17.3rad{s}^{-1}$
  3. $12.4rad{s}^{-1}$
  4. $9.82rad{s}^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of angular momentum about the center of the disc: L_initial = m*v*r = 0.02 * 5 * 0.1 = 0.01 kg m^2/s. L_final = (I_disc + m*r^2) * omega = (0.5 * 0.1 * 0.1^2 + 0.02 * 0.1^2) * omega = (0.0005 + 0.0002) * omega = 0.0007 * omega. Omega = 0.01 / 0.0007 = 14.2857 rad/s.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

A rod of length on two metal pads of same height from a height $h$. The coefficients of restitution of the metal pads are ${e} _{1}$ and ${e} _{2}$ (${e} _{1}> {e} _{2}$). The angular velocity of the rod after it recoils is

  1. $\cfrac { { e } _{ 1 } }{ { e } _{ 2 } } l\sqrt { 2gh } $
  2. $\cfrac { { e } _{ 1 }-{ e } _{ 2 } }{ l } \sqrt { 2gh } $
  3. $\cfrac { { e } _{ 1 }+1 }{ { e } _{ 2 }+1 } \sqrt { 2gh } $
  4. $\cfrac { { e } _{ 1 }+1 }{ { e } _{ 2 }-1 } \sqrt { 2gh } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A uniform rod AB of length $L$ and mass $M$ is lying on a smooth table. A small particle of mass $m$ strike the rod with a velocity $v _0$ at point C a distance x from the centre O. The particle comes to rest after collision. The value of $x$, so that point A of the rod remains stationary just after the collision,  is:

  1. $L/3$
  2. $L/6$
  3. $L/4$
  4. $L/12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For point A to be stationary, the velocity of the center of mass v_cm and the angular velocity omega must satisfy v_A = v_cm - omega * (L/2) = 0. Using conservation of linear and angular momentum: mv0 = M*v_cm and mv0*x = I*omega = (ML^2/12)*omega. Substituting v_cm = mv0/M and omega = 12mv0x/ML^2 into the condition gives x = L/6.

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$. When it moves from one point to a diametrically opposite point, its:

  1. momentum does not change
  2. momentum changes by $2Mv$
  3. $KE$ changes by $Mv^{2}$

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

Initial momentum vector is Mv at an angle. Final momentum vector at the diametrically opposite point is -Mv. The change in momentum is final - initial = -Mv - Mv = -2Mv. The magnitude of the change is 2Mv.

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

A particle starts from a point $z _0= I + i$, where $i
=\sqrt{-1}$ It moves horizontally away from origin by $2$ units and then
vertically away from origin by $3$ units to reach a point$ z _1$. From $z _1$
particle moves $\sqrt{5}$ units in the direction of $2\hat i + \hat j$ and
then it moves through an angle of $\cos e{c^{ - 1}}\sqrt 2 $ in anticlockwise
direction of a circle with centre at origin to reach a point $z _2$ . The arg $z _2$ is given by

  1. ${\sec ^{ - 1}}2$
  2. ${\cot ^{ - 1}}0$
  3. ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)$
  4. ${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The problem involves complex number transformations. Given the starting point and movements, the final argument calculation leads to an angle of 90 degrees, which corresponds to cot inverse of 0.

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 sphere of mass 0.5 kg and diameter 1 m rolls without sliding with a constant velocity of 5 m/s, the ratio of the rotational K.E. to the total kinetic energy of the sphere is :

  1. $\cfrac{7}{10}$
  2. $\cfrac{4}{9}$
  3. $\cfrac{2}{7}$
  4. $\cfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total kinetic energy for a rolling body is the sum of translational and rotational kinetic energy, given by K.E.(total) = (1/2)mv^2 + (1/2)Iomega^2. For a solid sphere, the moment of inertia I = (2/5)mr^2 and v = r omega. Substituting this in gives K.E.(rotational) = (1/5)mv^2 and K.E.(total) = (7/10)mv^2, leading to a ratio of 2/7.

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

If Kinetic energy is expressed as $mv^2/2$ for a particle undergoing uniform velocity motion, How is the kinetic energy expressed in case of the same particle, if it was rotating:

  1. $m \omega^2/2$
  2. $I \omega^2/2$
  3. $m V^2/2$
  4. $I V^2/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In rotation, m is replaced by I and v by  $\omega$. Thus the expression for kinetic energy becomes $I \omega^2/2$

The correct option is thus option (b)