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 physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A point sized sphere of mass $'m'$ is suspended from a point using a string of length $'l'$. It is pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is:

  1. $ ml\sqrt { gl } $
  2. $ ml\sqrt { 2gl } $
  3. $ml\sqrt { \dfrac { gl }{ 2 } } $
  4. $ ml\sqrt { 3gl } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By conservation of energy

$\dfrac{1}{2}mv^2=mgl$
$v=\sqrt{2gl}$
Angular momentum$=mvr=$$ml\sqrt{2gl}$

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

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

  1. $14.29\ rad\ s _1$
  2. $17.3\ rad\ s\ _1$
  3. $12.4\ rad\ s\ _1$
  4. $9.82\ rad\ s _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In this case, the angular momentum of  bit of putty about the axis of rotation = angular momentum of system of disc and bit of putty about the axis of rotation.


Let:

$M$ = Mass of puty

$m$ = Mass of disc

$ \therefore MvR=\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)\omega  $

 $ \omega =\dfrac{MvR}{\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)} $

 Putting all the values

 $ \omega =14.298\ m/s $

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A particle of mass $5kg$ is moving with a uniform speed $3\sqrt{2}$ in $XOY$ plane along the line $Y=X+4$. The magnitude of its angular momentum about the origin is:

  1. $40$units
  2. $60$units
  3. $0$
  4. $40\sqrt{2}$ units
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is identical to question 478324. For line Y = X + 4, perpendicular distance from origin is 4/√2 = 2√2. L = mvr_perpendicular = 5 × 3√2 × 2√2 = 60 units.

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A circular platform is mounted on a vertical frictionless axle. Its radius is $r=2m$ and its moment inertia is $I=200kg$ ${m}^{2}$. It is initially at rest. A $70kg$ man stands on the edge of the platform and begins to walk along the edge at speed ${v} _{0}=10{ms}^{-1}$ relative to the ground. The angular velocity of the platform is

  1. $1.2rad$ ${s}^{-1}$
  2. $0.4rad$ ${s}^{-1}$
  3. $2.0rad$ ${s}^{-1}$
  4. $7.0rad$ ${s}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As there is no external torque, thus $L$ is conserved.

Let angular velocity of the platform be $w.$
$L _i= L _f$
$0 = Iw - mv _o r$
$0 = 200 (w) - 70 (10 ) (2)$
$\implies w= 7    rad/s$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The blades of a windmill sweep out a circle of area $A$. If the wind flows at a velocity $v$ perpendicular to the circle, then the mass of the air of density $\rho$ passing through it in time $t$ is:

  1. $Av\rho t$
  2. $2Av\rho t$
  3. $Av^{2}\rho t$
  4. $\dfrac {1}{2}Av\rho t$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of wind flowing per second $= Av$
Mass of wind flowing per second $= Av\rho$
Mass of air passing in time $t\ s = Av\rho t$.

Multiple choice physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

A homogeneous disc of mass 2 kg and radius 15 cm is rotating about its axis (which is fixed) with an angular velocity of 4 rad/s. The linear momentum of the disc is :

  1. 1.2 kg-m/s

  2. 1.0 kg-m/s

  3. 0.6 kg-m/s

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The formula for angular momentum is:
$L=Iw$, where is the angular velocity and $I$  is the moment of inertia.
For a disk$,$
$I = \dfrac{1}{2}M{R^2},$ where $M$ is the mass of the disk and $R$ is the radius of the disk.
For this disk, 
$I = \frac{1}{2}M{R^2} = \dfrac{1}{2}\left( 2 \right) \times {\left( {0.15} \right)^2} = 0.0225\,kg\,{m^2}$
$L = Iw = 0.0225\,kg\,{m^2} \times \left( {4\,rad/s} \right) = 0.09\,kg\,{m^2}\,rad/s$
The linear momentum is related to the the angular momentum  by the formula; $L=pr.$
We can solve for p, the linear momentum in this equation to get,
$p = \dfrac{L}{r} = \dfrac{{0.09}}{{0.15}} = 0.6kgs$
Hence,
option $(C)$ is correct answer.
Multiple choice

For uniform circular motion, which values remain constant?

  1. orbital velocity and orbital speed

  2. centripetal acceleration and orbital speed

  3. centripetal acceleration and velocity

  4. velocity and centripetal force
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In uniform circular motion, the speed remains constant, and the magnitude of the centripetal acceleration remains constant, even though the direction of velocity and acceleration is constantly changing.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A thin uniform rod of mass $m$ and length $l$ is hinged at the lower end of a level floor and stands vertically. It is now allowed to fall, then its upper and will strike the floor with a velocity given by(A)$\sqrt { mgl }$(B) $\sqrt { 3gl }$(c)$\sqrt { 5gl }$ (D) $\sqrt { 2gl }$  Sol. 

  1. A

  2. B

  3. C

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

Using conservation of energy, the potential energy of the rod (mg*l/2) is converted into rotational kinetic energy (1/2 * I * omega^2). For a rod hinged at one end, I = ml^2/3. Thus, mg*l/2 = (1/2)*(ml^2/3)*omega^2. Solving for omega gives omega = sqrt(3g/l). The velocity of the tip is v = omega*l = sqrt(3gl).

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

A circular disc of mass $m$ and radius $r$ is rolling about its axis with a constant speed $v$. Its kinetic energy is 

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

Total kinetic energy = translational KE + rotational KE. For a rolling disc, K = (1/2)mv^2 + (1/2)I(omega^2). With I = (1/2)mr^2 and v = r*omega, K = (1/2)mv^2 + (1/2)(1/2)mr^2(v^2/r^2) = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.

Multiple choice position of point wrt ellipse ellipse maths

A rod of length $l$ rests against a vertical wall and a floor of a room.Let P be a point on the rod,nearer to its end on the wall, that divides its length in the ratio 1:2 if the rod begins to slide on the floor,then the locus of P is:

  1. an ellipse of eccentricity $\dfrac { 1 }{ 2 }$
  2. an ellipse of eccentricity $\dfrac { \sqrt { 3 } }{ 2 }$
  3. a circle of radius $\dfrac { l }{ 2 }$
  4. a circle of radius $\dfrac { \sqrt { 3 } }{ 2 } l$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the rod be AB of length l, with A on the y-axis and B on the x-axis. A point P(x,y) dividing the rod in ratio 1:2 has coordinates x = (2*0 + 1*x_B)/3 and y = (2*y_A + 1*0)/3. Thus x = x_B/3 and y = 2y_A/3. Since x_B^2 + y_A^2 = l^2, we have (3x)^2 + (3y/2)^2 = l^2, which is 9x^2 + 9y^2/4 = l^2. This is an ellipse with semi-axes a = l/3 and b = 2l/3. Eccentricity e = sqrt(1 - (l/3)^2/(2l/3)^2) = sqrt(1 - 1/4) = sqrt(3)/2.

Multiple choice

What is the scientific term for the spinning motion performed by breakdancers?

  1. Angular momentum

  2. Centripetal force

  3. Rotational inertia

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

Angular momentum is a measure of the amount of rotation of an object around an axis. In breakdancing, angular momentum is generated by the dancer's body spinning around its center of mass.

Multiple choice

The gyroscopic effect is a phenomenon that occurs when a spinning object experiences a:

  1. Precession

  2. Nutation

  3. Both precession and nutation
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The gyroscopic effect is a phenomenon that occurs when a spinning object experiences a torque that is perpendicular to its axis of rotation. This torque causes the axis of rotation to precess, or wobble, around the direction of the applied torque. Additionally, the spinning object may also experience nutation, which is a small, periodic oscillation of the axis of rotation around the direction of precession.

Multiple choice

What is the relationship between torque and angular acceleration?

  1. Torque = Angular Acceleration x Moment of Inertia

  2. Torque = Angular Acceleration / Moment of Inertia

  3. Torque = Angular Acceleration + Moment of Inertia

  4. Torque = Angular Acceleration - Moment of Inertia
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Torque is the product of force and the perpendicular distance from the axis of rotation. According to Newton's second law of motion for rotational motion, torque is equal to the moment of inertia of the object multiplied by its angular acceleration.

Multiple choice

In a role-playing game, a player character swings a sword with a mass of 2 kg at a speed of 10 m/s. What is the angular momentum of the sword?

  1. 20 kg*m^2/s

  2. 30 kg*m^2/s

  3. 40 kg*m^2/s

  4. 50 kg*m^2/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Angular momentum (L) is given by the equation: L = I * ω, where I is the moment of inertia and ω is the angular velocity. Assuming the sword is a thin rod rotating about its center of mass, the moment of inertia is given by: I = (1/12) * m * L^2, where m is the mass and L is the length of the sword. Since the length of the sword is not given, we cannot calculate the exact value of the angular momentum. However, we can say that the angular momentum is proportional to the mass and the square of the velocity.