A bob of mass M is suspended through a massless string of length L. The horizontal velocity at position A is sufficient enough to make it reach at point B. The angle at which the speed of bob is half of that at A, satisfies
A
θ=π4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π4<θ<π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π2<θ<3π4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3π4<θ<π
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D3π4<θ<π The bob attached to string just reaches at B that means v=0 there, hence for just completing the circular loop, velocity at bottom most point: v=√5gL....(i)
At some position C string makes angle θ with vertical and velocity of bob is v2.
Height above bottom point A is h at position C h=L−Lcosθ=L(1−cosθ)...(ii)
If an object is rising along the vertical circular path with initial velocity u, then at height h by applying mechanical energy conservation final velocity v is given by: 12mu2=12mv2+mgh v2=u2−2gh...(iii)
for position A to C u=v,v=v2
From Eq. (ii) and (iii) ⇒(v2)2=v2−2gL(1−cosθ)
⇒3v24=2gL(1−cosθ)
Putting value of v from Eq. (i), 15gL4=2gL(1−cosθ) 158=1−cosθ ∴cosθ=−78=−0.875
It indicates that value of cosθ is −ve and close to −1, so θ must be an angle of 2nd quadrant which is near to 180∘orπ ∴ Hence range (3π4<θ<π) suits most.