1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Consider two masses with m1>m2 connected by a light inextensible string that passes over a pulley of radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance 2h. When the two masses pass each other, the speed of the masses is proportional to.

A
 m1m2m1+m2+IR2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
 (m1+m2)(m1m2)m1+m2+IR2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
 m1+m2+IR2m1m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
 IR2m1+m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A  ⎷m1−m2m1+m2+IR2The objects meet when each has travelled distance h The lost PE=(m1−m2)gh This equals the gain in KE, so KE=(m1−m2)gh=12(m1+m2)v2+12I(VR)2 Assuming that Rwas given in your diagram as the radius of the pulley. (m1−m2)gh=12(m1+m2+I/R2)v2v=√2(m1−m2)ghm1+m2+I/R2vα  ⎷m1−m2m1+m2+IR2 The objects meet when each has travelled distance h The lost PE=(m1−m2)gh This equals the gain in KE, so KE=(m1−m2)gh=12(m1+m2)v2+12I(VR)2 Assuming that Rwas given in your diagram as the radius of the pulley. (m1−m2)gh=12(m1+m2+I/R2)v2v=√2(m1−m2)ghm1+m2+I/R2vα  ⎷m1−m2m1+m2+IR2

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Work Energy and Power
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program