Physics

Rotational Mechanics and Machines

91 Questions

Rotational mechanics involves the study of angular motion, torque, gears, and rotational energy. These concepts are frequently tested in physics sections across various competitive assessments. Practicing these problems builds a strong foundation in machine kinematics and dynamics.

angular kinematicsgears and speedrotational energytorque calculationsmoment of inertia

Rotational Mechanics and Machines Questions

Multiple choice

A planetary gear train has four gears and one carrier. Angular velocities of the gears are $\infty_1$,$\infty_4$,$\infty_4$ and$\infty_4$ respectively. The carrier rotates with angular velocity $\infty_4$.

What is the relation between the angular velocities of Gear 1 and Gear 4?
  1. $\begin{matrix} \ \omega_1 -\omega_5 \\ \ \omega_1 -\omega_5 \\ \end{matrix}=6$
  2. $\frac{\omega_1-\omega_5}{\omega_1\hspace{0.3cm}\omega_5}=6$
  3. $\frac{\omega_1-\omega_5}{\omega_4-\omega_5}=-\bigg(\frac{2}{3}\bigg)$
  4. $\frac{\omega_1-\omega_5}{\omega_4-\omega_5}=\frac{8}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\frac{\omega_1-\omega_5}{\omega_4-\omega_5}=\frac{x}{x\times\frac{Z_1Z_3}{Z_2Z_4}}=\frac{Z_2Z_4}{Z_1Z_3}\\ \frac{\omega_1-\omega_5}{\omega_4-\omega_5}=\frac{45\times40}{15\times20}=3\times2=6$

Multiple choice

A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1 m. Assuming that the wheel and the ground are both rigid and that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately

  1. zero
  2. $\frac{1}{3}rad/s$
  3. $\sqrt{\frac{10}{3}}red/s$
  4. $\frac{10}{3}red/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

An instantaneous configuration of a four-bar mechanism, whose plane is horizontal, is shown in the figure below. At this instant, the angular velocity and angular acceleration of link O2A are $\omega8red/s$ and$\alpha=0$, respectively, and the driving torque $(t)$ is zero. The link O2A is balanced so that its center of mass falls at O2.

At the instant considered, what is the magnitude of the angular velocity of O4B?

  1. 1 rad/s

  2. 3 rad/s

  3. 8 rad/s
  4. $\frac{64}{3}red/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{Let,w_4 is the angular velocity of link $O_4$B}\\ \text{From the triangle ABC,}\\ \hspace{2cm}tan\theta=\frac{100}{240}=\frac{5}{12}\hspace{6cm}...(i)\\ \theta=tan^{-1}(\frac{5}{12})=22.62^0 \\ \text{Also from the triangleO_1O_2A,}$

Multiple choice

A band brake consists of a lever attached to one end of the band. The other end of the band is fixed to the ground. The wheel has a radius of 200 mm and the wrap angle of the band is 270°. The braking force applied to the lever is limited to 100N, and the coefficient of friction between the band and the wheel is 0.5. No other information is given.

The maximum tension that can be generated in the band during braking is

  1. 1200 N

  2. 2110 N

  3. 3224 N

  4. 4420 N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

A band brake consists of a lever attached to one end of the band. The other end of the band is fixed to the ground. The wheel has a radius of 200 mm and the wrap angle of the band is 270°. The braking force applied to the lever is limited to 100 N, and the coefficient of friction between the band and the wheel is 0.5. No other information is given.

The maximum wheel torque that can be completely braked is

  1. 200 N.m

  2. 382 N.m

  3. 604 N.m

  4. 844 N.m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice gear torque

Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 1200 rpm and transmits a torque of 20 Nm. The torque transmitted by gear is

  1. 7.0 Nm

  2. 20 Nm

  3. 35 Nm

  4. 60 Nm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The gear ratio is the ratio of the number of teeth on the gear to the pinion (120/40 = 3). Since torque is proportional to the gear ratio, the torque on the gear is 3 times the torque on the pinion (20 Nm * 3 = 60 Nm).

Multiple choice mathematics and statistics angle and its measurement directed angles

If a bicycle wheel has 48 spokes the angle between the adjacent pair spokes is :

  1. $\left ( 6\frac{1}{2} \right )^0$
  2. $\left ( 7\frac{1}{2} \right )^0$
  3. $\left ( 7\frac{1}{3} \right )^0$
  4. $\left ( 6\frac{2}{3} \right )^0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

required angle = $\frac{360}{48}$
$\frac{30}{4}$ = $\frac{15}{2}$= $\left ( 7\frac{1}{2} \right )^0$

Multiple choice motional emf physics

The axle of a circular wheel of radius R is held horizontally by two identical strings of equal lengths separated by a distance D. The tension in each string is $T _0$. The rim of the wheel carries a total charge $+$Q distributed uniformly on it. The wheel is vertical and is kept in a uniform vertical magnetic field $\vec{B}$. It is now rotated at an angular speed $\omega$. If the string break at a tension of $3T _0/2$, than the maximum possible value of $\omega$ at which the wheel can be rotated without breaking a string is $\dfrac{DT _0}{QBR^2}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

A wheel of radius $1$ meters rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:

  1. $2 \pi$
  2. $\sqrt 2 \pi$
  3. $\sqrt {{\pi ^2} + 4} $
  4. $\pi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let fro $A$ to $B$, where complete half a rotation.
we have,
$PQ=\alpha =$ half circumference $=\pi \ m$
and $P' Q=$ diameter $=2\ m$
so we have displacement of $P$
$x=|PP;|=\sqrt {PQ^2 +P; Q^2}$ (Pythagoras theorem)
$\Rightarrow \ x=\sqrt {\pi^2 +4}m$
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

What is the displacement of the point on the wheel initially in contact with the ground when the wheel rolls forwards half of revolution? Take the radius of the wheel as $'R'$ and the x-axis in the forward direction

  1. $R\sqrt{\pi^2 + 9}, Tan^{-1}\left(\dfrac{3}{\pi}\right)$ with x-axis
  2. $R\sqrt{\pi^2 + 4}$ and angle $Tan^{-1}\left(\dfrac{2}{\pi}\right)$ with x-axis
  3. $R\sqrt{\pi^2 + 16}, Tan^{-1}\left(\dfrac{4}{\pi}\right)$ with x-axis

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For half a revolution, the center moves forward by pi*R. The point initially at the bottom moves to the top, so its vertical displacement is 2R. The total displacement is sqrt((pi*R)^2 + (2R)^2) = R*sqrt(pi^2 + 4). The angle is tan^-1(vertical/horizontal) = tan^-1(2R/pi*R) = tan^-1(2/pi).

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 whose radius is $r$ and moment of inertia about its-own axis is $I$, can rotate freely about its own horizontal axis. A rope is wrapped on the wheel. A boy of mass $m$ is suspended from the free end of the rope. The body is released from rest. The velocity of the body after falling a distance $h$ would be- 

  1. $\left(\dfrac{mgh}{I}\right)^{{1}/{2}}$
  2. $\left(\dfrac{2mgh}{m+I}^{{1}/{2}}\right)$
  3. $\left(\dfrac{2mgh}{m+I/r^2}\right)^{{1}/{2}}$
  4. $\left(\dfrac{m +I}{mgh}\right)^{{1}/{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of energy: mgh = (1/2)mv^2 + (1/2)I(omega)^2. Since v = omega*r, omega = v/r. So mgh = (1/2)mv^2 + (1/2)I(v/r)^2. Solving for v gives v = sqrt(2mgh / (m + I/r^2)).

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 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 flywheel is in the form of a solid circular wheel of mass 72 kg and radius 50cm and it makes 70 r.p.m. then the energy of revolution is:

  1. 245534 J

  2. 24000 J

  3. 4795000J

  4. 4791600 J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$K.E=\cfrac{1}{2}mv^2\rightarrow(1)\v=r\omega$

Put in $(1)$
$K.E=\cfrac{1}{2}mr^2\omega^2$
Given data,
$m=72kg\r=50cm\ \omega=70rev/min$
$1rev=2\pi rad\1min=60sec\ \omega=\cfrac{70\times2\pi}{60}=2.33\times3.14\ \omega=7.3rad/sec$
So, $K.E=\cfrac{1}{2}mr^2\omega^2\Rightarrow\cfrac{1}{2}\times72\times50\times50\times\cfrac{7.3}{10}\times\cfrac{7.3}{10}\ K.E=4791600J$