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A mass m(1) lies on fixed, smooth cylind...

A mass `m_(1)` lies on fixed, smooth cylinder . An ideal cord attached to `m_(1)` passes over the cylinder and is connected to mass `m_(2)` as shown in the figure.
`(a)` Find the value of `theta (` shown in diagram `)` for which the system is in equilibrium.
`(b)` Given `m_(1)=5 kg, m_(2)=4kg` . The system is released from rest when `theta =30^(@)` . Find the magnitude of acceleration of mass `m_(1)` just after the system is released.

Text Solution

Verified by Experts

The correct Answer is:
`sin theta=(m_(2))/(m_(1))`
`(b) (15)/(9)m//s^(2)`

`(a)` The system is in equilibrium when `m_(1)g sin theta =m_(1)g`
or `sin theta =(m_(2))/(m_(1))`
`(b)` Let the tangential acceleration of `m_(1)` be `a`.
`:. m_(2)g-m_(1)g sin theta=(m_(1)+m_(2))a`
`a=(40-25)/(9)=(15)/(9)m//s^(2)`
the normal acceleration of `m_(1)` is zero.
`:'` speed of `m_(1)` is zero.
`:. ` The magnitude of acceleration of `m_(1)=(5)/(3)m//s^(2)`
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