A bob of mass m=50g is suspended form the ceiling of a trolley by a light inextensible string. If the trolley accelerates horizontally, the string makes an angle `thata=37^(@)` with the vertical. Find the acceleration of the trolley.

Options

A. `10 m//s^(2)`
B. `12.5 m//s^(2)`
C. `7.5 m//s^(2)`
D. `7.5 m//s`

Answer

C

Explanation

Method 1: Problem solving from ground frame. The observer on the ground will observe the bob moving with acceleration a same as trolley. In FBD, he will observe only two real forces, weight and tension force.
The bob is at equilbrium in vertical direction, hence writing equation of motion in vertical direction
`T cos theta =mg` ...(i)
along horizontal direction the bob is accelerating with acceleration a.
`T sin thata =ma` ...(ii) Using (i) and (ii) `tan theta=a//g`
`:. a=g tan theta = 10 tan 37^(@)=7.5 ms^(2)`
Therefore, acceleration of the trolley is `7.5 ms^(-2)`
` `
Method 2: Problem solving from non-inertial frame. If an observer sitting in the trolley and observes the bob, he will observe the bob at rest. If he asks to draw free-body diagram of bob he will apply pseudo force apart from real forces action on the bob. The bob is at equilibrium in vertical direction, hence writting equation of motion in vertical direction
`T cos theta = mg`..(iii)
Along horizontal direction the bob is at rest
`T sin theta =ma` ...(iv)
Using (iii) and (iv) `tan theta=a//g`
`:. a=g tan theta = 10 tan 37^(@)=7.5 ms^(2)`