Question

# A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude 0.5N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rads−1 is:

A
2
B
4
C
5
D
7
Solution
Verified by Toppr

#### The torque acting on each of the points X, Y and Z are in anticlockwise direction and equal in magnitude.τtotal=3FRsin30o=3FR2Moment of inertia I=MR22∴τtotal=Iα=MR22α=3FR2∴1.5×(0.52)2α=3×0.5×0.52⇒α=2From rotational kinematics equationω=ω0+αtAt t=1s∴ω=2t=2×1=2rad/s

1
Similar Questions
Q1

A disc having mass 1 kg and radius 2 m is pivoted about its centre. Initially, it is at rest. A force F=(5t2+9) N starts acting tangentially on it. Find the angular velocity of the disc after 3 seconds in rad/s

View Solution
Q2

A disc having mass 1 kg and radius 2 m is pivoted about its centre. Initially, it is at rest. A force F=(5t2+9) N starts acting tangentially on it. Find the angular velocity of the disc after 3 seconds in rad/s

View Solution
Q3

A uniform ring and a uniform disc of same mass and radius are rotating with same angular speed. If a constant force is applied tangentially on the ring in opposite direction, it stops after n revolution. If the same force is applied on disc in opposing sense tangentially then the number of revolution made by it before it stops, is:

View Solution
Q4

A uniform disc of mass M and radius R is hinged at its centre C. A force F is applied on the disc as shown. At this instant, angular acceleration of the disc is:

View Solution
Q5

A circular disc is rotating about its own axis. An external opposing torque of magnitude 0.02Nm is applied on the disc, due to which it comes to rest in 5 seconds. The initial angular momentum of the disc is:

View Solution
Solve
Guides