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

# A small bob hangs at one end of a massless string of length l and the other end of the string is fixed. The bob is given a sharp hit to impart it a horizontal speed of √3gl. Find the angle (θ) made by the string with the vertical when the string becomes slack.

A
θ=cos1(23)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
θ=cos1(34)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
θ=cos1(13)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
θ=cos1(53)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C θ=cos−1(13)Horizontal speed of the bob at point A is, u=√3gl Let m be the mass of the bob. At point B, string becomes slack i.e (T=0) in the string & angle made by the string with vertical is θ. Let speed of the bob be v Applying Newton's 2nd law towards the centre of the vertical circle at point B: mgcosθ+T=mv2l ...(i) (putting T=0 in Eq. i) ⇒mgcosθ=mv2l ∴v2=lgcosθ ....(ii) Height to which bob rises above point A h=lcosθ+l=l(1+cosθ) ...(iii) Applying mechanical energy conservation from A to B: ⇒Loss in KE=Gain in PE 12mu2−12mv2=mgh ⇒v2=u2−2gh ...(iv) From Eq. (ii), (iii), & (iv), lgcosθ=3gl−2gl−2glcosθ ⇒3glcosθ=gl ⇒cosθ=13 ∴θ=cos−1(13)

Suggest Corrections
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program