Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A uniform circular disc of mass ‘m’ and radius ‘a’ can rotate in a vertical plane about a horizontal axis through a point O on its circumference. The disc is held with the diameter OA through O horizontal and then released. If a constant frictional couple of 'mga/2pie' opposes the motion, the angular speed when OA is vertical will be


A uniform circular disc of mass ‘m’ and radius ‘a’ can rotate in a vertical plane about a horizontal axis through a point O on its circumference. The disc is held with the diameter OA through O horizontal and then released. If a constant frictional couple of 'mga/2pie' opposes the motion, the angular speed when OA is vertical will be 


Grade:12

1 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

I=ma2/2 +ma2=(3/2)ma^2 , loss of potential energy=mga, work done by frictional couple=(mga/2pie)(pie/2), therefore

if w is angular speed, Iw^2= mga-mga/4=3mga/4

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free