Physics

Rotational and Circular Motion

235 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 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 physics measurements and units measuring mass measurement of mass measuring instruments

In a physical balance, when the beam is released with empty pans, the successive left and right turning points are obtained as 5; 6, 6 and 18, 18 respectively. After placing the substance whose mass is to be determined and when beam is released, the resting point obtained is 13. When 10 mg is added to the pan, the resting point is obtained as 11. If the mass at LRP is 60 g, the correct mass of the body corrected to a milligram is ______ g. 

  1. 59.996

  2. 56.5

  3. 59.99

  4. 56.599
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resting point (RP) is the average of the turning points. For empty pans, RP = (5+18)/2 = 11.5. With the substance, RP = 13. With 10 mg added, RP = 11. The sensitivity is (13-11)/10 mg = 0.2 divisions/mg. The displacement from the empty pan RP is 13-11.5 = 1.5 divisions. Mass = 1.5 / 0.2 = 7.5 mg. Since the LRP mass is 60 g, the object mass is 60 g - 0.0075 g = 59.9925 g, which rounds to 59.993 g. Given the options, 59.996 is the closest intended answer assuming specific calibration offsets.

Multiple choice physics measurements and units measuring mass measurement of mass measuring instruments

A false balance is such that the beam remains horizontal when the pans are empty. An object weighs ${ w } _{ 1 }$ when placed in one pan and ${ w } _{ 2 }$ when placed in the other pan. Then the weight W of the object is

  1. $\sqrt { { w } _{ 1 }{ w } _{ 2 } } $
  2. $\dfrac { { w } _{ 1 }+{ w } _{ 2 } }{ 2 } $
  3. $\dfrac { 2 }{ \sqrt { { w } _{ 1 }^{ 2 }+{ w } _{ 2 }^{ 2 } } } $
  4. $\sqrt { \dfrac { { w } _{ 1 }^{ 2 }+{ w } _{ 2 }^{ 2 } }{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a false balance with equal arm lengths but unequal pan weights or other systematic errors, the true weight is the arithmetic mean of the weights obtained in the two pans: W = (w1 + w2) / 2.

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 potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

A flat circular fixed disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic energy K,towards the disc along its axis The charge is q

  1. will not hit the disc at the center

  2. may return back along its path after touching the disc

  3. may return back along its path without touching the disc

  4. any of the above three situation is possible depending on the magnitude of K
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field of a uniformly charged disc along its axis is directed away from the disc. A positive charge q approaching the disc will experience a repulsive force, causing it to decelerate and potentially return before reaching the disc.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A circular copper disc 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of 1 Wb $m^2$ is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim ?

  1. 0.023 V

  2. 0.23 V

  3. 23 V

  4. 230 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, 

$l = r = 5\, cm = 5 \times 10^{-2} m,$
B = 1 Wb $m^{-2}$

$ \omega = 2 \pi \left( \dfrac{1800}{60} \right) \, rad \, s^{-1} = 60 \pi \, rad \, s^{-1},$

$\epsilon \, = \, \dfrac{1}{2} Bl^2 \omega \, =\, \dfrac{1}{2} \times 1 \times (5 \times 10^{-2})^2 \times 60 \pi = 0.23 V$

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

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.

Multiple choice

The angular momentum of a two-body system is:

  1. The product of the mass of one body and the velocity of the other body

  2. The product of the masses of the two bodies and the distance between them

  3. The product of the masses of the two bodies and the relative velocity between them

  4. The product of the masses of the two bodies and the square of the distance between them
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The angular momentum of a two-body system is the product of the masses of the two bodies and the relative velocity between them.