Q 1 :    

A string is wrapped around the rim of a wheel of moment of inertia 0.40 km2 and radius 10 cm. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of 40 N. The angular velocity of the wheel after 10 s is x rad/s, where x is _______ .            [2024]



(100)      τ=FR=Iα=40×0.1=0.4α

               α=40.4=10rad/s2

               ωf=ω0+αt=0+10×10=100rad/sec.

 



Q 2 :    

A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5t28t, where θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t = 2 s?          [2025]

  • 60MR2

     

  • 72MR2

     

  • 108MR2

     

  • 8MR2

     

(1)

Given: θ=5t28t

ω=dθdt=10t8

α=dωdt=10 rads2

Power, P=τω=(Iα)ω

 P=(MR22)αω=(MR22)(10)(10t8)

Put t = 2 s, P=60MR2



Q 3 :    

 

A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of 2 rad/s2. Moment of inertia of the wheel is _____ kgm2. (Acceleration due to gravity = 10 m/s2)          [2025]



(1)

τ=FR=Iα

 I=FRα=10×0.22= 1 kgm2



Q 4 :    

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 πNm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is __________ m.          [2025]



(40)

Given m = 1 kg

ωi=1800 rpm=1800×2π60=60πradsec

ωf=2100 rpm=2100×2π60=70πradsec

τext=25πNm, t=40 sec

Using equation of motion, ωf=ωi+αt

70π=60π+α(40)  α=π4rad/sec2

Also, τ=Iα τ=mR24α 

25π=1×R24×π4  R=20 m

Hence, diameter of disk = 2R = 2 × 20 = 40 m