Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Apne doubts clear karein ab Whatsapp par bhi. Try it now.

CLICK HERE

Watch 1000+ concepts & tricky questions explained!

100+

7

Text Solution

Solution :

Here M = kg, R = 40 cm = 0.4 m <br> Moment of inertia of the hollow cylinder about its axis <br> `I=MR^(2)=3(0.4)^(2)=0.48kgm^(2)` <br> Force applied F = 30 N <br> `therefore` Torque, `tau=FxxR=30xx0.4=12N-m`. <br> If `alpha` is the angular acceleration, produced, then from `tau=lalpha` <br> `alpha=(tau)/(l)=(12)/(0.48)=25" rads"^(-2)` <br> Linear acceleration, `a=Ralpha=0.4xx25` <br> `=10m//s^(2)`.