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Chapter 10: Q87P (page 293)

In Fig.10-55, a wheel of radius 0.20mis mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a2.0Kgbox that slides on a frictionless surface inclined at angle 胃=20掳with the horizontal. The box accelerates down the surface at2.0ms2. What is the rotational inertia of the wheel about the axle?

Short Answer

Expert verified

Rotational inertia about wheel is0.054kgm2

Step by step solution

01

Given

  1. Radius of wheel is0.2m
  2. Mass of box is2.0kg
  3. Acceleration is2.0m/s2
  4. Angle of inclination is20掳
02

To understand the concept

Use Newton鈥檚 second law to find tension. Using the formula for the torque in terms of tension and radius and also using the formula in terms of moment of inertia and angular acceleration, find the acceleration.

Formula:

Fnet=M脳a蟿=I伪=T脳r伪=ar

03

Draw the Free Body Diagram

04

Calculate the rotational inertia of the wheel about the axle

Net force acting on block is as follow

Fnet=magsin胃-TA

Here胃=20掳

鈬�Fnet=magsin20-TA

According to Newton鈥檚 law,

Fnet=maa

So

maa=magsin20-TA鈬�2脳2=2脳9.8脳sin20-TA鈬�TA=2.7N

Here torque is due to tension so

蟿=r脳T=I脳伪鈬�r脳TA=I脳ar鈬�I=TA脳r2a鈬�I=2.7脳0.222鈬�I=0.054kgm2

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