Physics

Rotational Mechanics and Machines

79 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 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, $K.E=\cfrac{1}{2}mr^2\omega^2\Rightarrow\cfrac{1}{2}\times72\times50\times50\times\cfrac{7.3}{10}\times\cfrac{7.3}{10}\ K.E=4791600J$

Multiple choice physics force and newton's laws of motion concept of inertia galileo's law and inertia mass and inertia

The sparks produced during sharpening of a knife against a grinding wheel leaves the rim of the wheel tangentially. This is due to which of the following?

  1. Inertia of rest

  2. Inertia of motion

  3. Inertia of direction

  4. Force applied
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Just after the sparks are produced, the force acting on the sparks becomes zero and due to inertia of direction, they start to move in their initial direction of motion which is tangential to the rim of the grinding wheel.

Multiple choice introduction to ratio and percentages comparing quantities maths

A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is

  1. 4

  2. 9

  3. 12

  4. 49
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
$\therefore 14 : 6 :: 21 : x \Leftrightarrow 14 \times x = 6 \times 21$
$\Rightarrow x = \dfrac{6 \times 21}{14}$
$\Rightarrow x = 9$

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of $15  N$ can be obtained at each of its brake shoe by exerting a force of $0.5  N$ on the pedal?

  1. $1: 60$
  2. $1: 30$
  3. $1: 15$
  4. $1: 45$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $P _{1} = P _{2}$

$\Rightarrow \dfrac{F _{1}}{A _{1}} = \dfrac{F _{2}}{A _{2}} $

$\Rightarrow \dfrac{A _{1}}{A _{2}} = \dfrac{F _{1}}{F _{2}}  = \dfrac{1}{30} $

$\Rightarrow  1:30 $

Multiple choice physics a little effort, lot of work levers and pulleys pulley pulleys

In a wheel-and-axle :

  1. Wheel rotates freely on axle

  2. Wheel is fixed to the axle

  3. Direction of the lbrce remains same

  4. Mechanical advantage is always more than one
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A wheel-and-axle is defined by the wheel and axle being fixed together so that they rotate as a single unit.

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

The diameter of a wheel is 98 cm The number of revolutions it will have to cover a distance of 1540 m is

  1. 500

  2. 600

  3. 700

  4. 800
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the diameter of wheel is 98 cm

Then radius of wheel =$\frac{98}{2}=49cm$
Then circumference of wheel =$2\times \frac{22}{7}\times 49=308cm$
Then the number of revolution in distance of 1540 m=$\frac{1540\times 100}{308}=\frac{154000}{308}=500$

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

The front wheels of a wagon are $2$ m in circumference and the back wheels are $3$ m feet in circumference. When the front wheels have made $10$ more revolutions than the back wheels, how many metres has the wagon travelled?

  1. $40$ feet
  2. $50$ feet
  3. $60$ feet
  4. $80$ feet
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the back wheels make n revolutions. Then front wheels make n+10 revolutions. Distance travelled by front = 2(n+10) and by back = 3n. Equating: 2(n+10) = 3n, so 2n+20 = 3n, giving n=20. Distance = 3*20 = 60 feet.

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

A bicycle wheel has diameter 1m. If the bicycle travels one kilometer, then the number of revolutions the wheel make is.

  1. $\dfrac {1}{\Pi }$
  2. $\dfrac {100}{\Pi }$
  3. $\dfrac {500}{\Pi }$
  4. $\dfrac {1000}{\Pi }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the number of revolution of the wheel is n.
Then,
n $\times$ circumference of wheel = Distance travelled by bicycle
$n \times  2\Pi  \times \frac {1}{2}=1$ kilometer 
$n \times  \Pi $=1000 meter
$n=\frac {1000}{\Pi }$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A cyclist riding with a speed of $27kmph$. As he approaches a circular turn on the road of radius $80m$, he applies brakes and reduces his speed at the constant rate of $0.50m/s$ every second. The net acceleration of the cyclist on the circular turn is 

  1. $0.5 \quad m/s^2$
  2. $0.86\quad m/s^2$
  3. $0.56 \quad m/s^2$
  4. $1 \quad m/s^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ v = 27 Km/hr$ 

$  = \cfrac{27 \times 1000}{3600} = 7.5 m/sec$
Centripetal acceleration = $ \cfrac{(7.5)^2}{80} = 0.7 m/s^2$
Tangential acceleration = $ -0.5m/s^2$ 
$ \therefore a _{net} = \sqrt{(0.7)^2 + (0.5)^2} = 0.86m/s^2$

Multiple choice physics types of energy renewable and non-renewable resources renewable and non-renewable sources of energy substances, objects and energy

If $484J$ of energy is spent in increasing the speed of a wheel from $60rpm$ to $360\ rpm$, the $M.I$ of the wheel is

  1. $1.6\ kg\ m^{2}$
  2. $0.3\ kg\ m^{2}$
  3. $0.7\ kg\ m^{2}$
  4. $1.2\ kg\ m^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { \eta _{ 1 } }=60rpm=\dfrac { { 60 } }{ { 60 } } rps=1rps \ \therefore { w _{ 1 } }=2r{ \eta _{ 1 } }=2\pi  \ { \eta _{ 2 } }=\dfrac { { 360 } }{ { 60 } } rpm=6rps \ { w _{ 2 } }=2\pi { \eta _{ 2 } }=12\pi  \end{array}$

484 joule of energy is spent in increasing.
$\begin{array}{l} \frac { 1 }{ 2 } Iw _{ 2 }^{ 2 }-\dfrac { 1 }{ 2 } Iw _{ 1 }^{ 2 }=484 \ \dfrac { I }{ 2 } \left( { { w _{ 2 } }+{ w _{ 1 } } } \right) \left( { { w _{ 2 } }-{ w _{ 1 } } } \right) =484 \ \dfrac { I }{ 2 } \left( { 12\pi +2\pi  } \right) \left( { 12\pi -2\pi  } \right) =484 \ I\, \, 14\pi .10\pi =484 \ I=\dfrac { { 484\times 2 } }{ { 140\times { \pi ^{ 2 } } } } =\dfrac { { 484\times 2 } }{ { 140\times 9.86 } }  \ =0.7\, \, kg\, \, { m^{ 2 } } \end{array}$
Option C.

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

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2t^{3}-6t^{2}$. The torque on the wheel becomes zero at :

  1. $t=1s$
  2. $t=0.5\ s$
  3. $t=0.25\ s$
  4. $t=2s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $\theta (t)=2t^3-6t^2$

We, know $T=\propto \dfrac{Id^2 \theta}{dt^2}=I\dfrac{d}{dt}(\dfrac{d \theta}{dt})=I\dfrac{d}{dt}(6t^2-12t)=I(12t-12)=0$

Then, $12t-12=0\Rightarrow t=1s$