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Fundamentals Of Physics
Found in: Page 293
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

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

Rotational inertia about wheel is 0.054 kgm2

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given

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

Step 2: To understand the concept

Use Newton’s 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×aτ=Iα=T×r α=ar

Step 3: Draw the Free Body Diagram

Step 4: Calculate the rotational inertia of the wheel about the axle

Net force acting on block is as follow

Fnet=magsinθ-TA

Hereθ=20°

⇒Fnet=magsin20-TA

According to Newton’s law,

Fnet=maa

So

maa=magsin20-TA⇒2×2=2×9.8×sin20-TA⇒TA=2.7 N

Here torque is due to tension so

τ=r×T=I×α⇒r×TA=I×ar⇒I=TA× r2a⇒I=2.7×0.222⇒I=0.054 kgm2

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