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

A uniform disc of mass m and radius r is pivoted at point p and is free to rotate in vertical plane. The centre C of disc is initially in horizontal position with p. if it is released from this position then it's angular acceleration when the line PC is inclined to the horizontal at an angle q

A
α=2gcosθ6R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
α=5gcosθ5R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
α=4gcosθ3R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
α=2gcosθ3R
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D α=2gcosθ3R
we need to find the torque on the disc about its point P when it is at angle with the horizontal.

now by the equation we can say
τ=F×R×sin(90θ)
plug in values in it
τ=mg×R×cosθ
now we can use
τ=Iα
now the moment of inertia of disc about point P
I=32mR2
now plug in all values in above equation
mgRcosθ=32mR2×a
so angular acceleration is given as
α=2gcosθ3R
so above is the angular acceleration of disc at given position
Hence,
option (D) is correct answer.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Moment of Inertia of Solid Bodies
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon