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

# A particle tied to the end of a string oscillates along a circular arc in a vertical plane. The other end of the string is fixed at the centre of the circle. If the string has a breaking strength of twice the weight of the particle; (x) Find the maximum distance that the particle can cover in one cycle of oscillation. The length of the string is 50 cm. (y) Find the tension in the extreme position (z) Find the net acceleration of the particle at extreme (i) 2π3m (ii)mg4 (iii) g√23 (iv)mg2 (v)π3m (vi) g√32

A

x - i, y - ii, z - iii

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

x - v, y - ii, z - iii

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

x - v, y - iv, z - iii

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

x - i, y - iv, z - vi

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D x - i, y - iv, z - vi As the maximum tension occurs at the lowest position, tension at the bottom can be at most 2mg (where m is the mass of particle). Considering Forces on the particle at the bottom: Where u : velocity at bottom 2mg - mg = mu2l⇒u=√lg (x) Let θ = angular amplitude from bottom to the extreme, loss in KE = Gain in GPE 12mu2=mg(l−lcosθ)⇒12mlg=mg(l−lcosθ) (1−cosθ)=12⇒θ=60∘ Length of are covered in one cycle = 4(lθ)=4(12π3)=2π3m (y) At extremes, the speed is zero and hence radial acceleration is zero. Balancing radial forces we get:TE=mgcosθ=mg2 (z) At extremes: ar=0m/s2 and at=Fim=mgsin60∘m=g√32 ⇒ Net acceleration = g√32

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
The Law of Conservation of Mechanical Energy
PHYSICS
Watch in App
Join BYJU'S Learning Program