Home
Class 11
PHYSICS
A flywheel rotating at a speed of 600 r...

A flywheel rotating at a speed of `600` rpm about its axis is brought to rest by applying a constant torque for `10` seconds. Find the angular deceleration and angular velocity `5` second after the application of the torque.

Text Solution

Verified by Experts

The correct Answer is:
A, B

A wheel rotating at a speed of 600 rpm
`omega_0=600 rpm`
=10 revolutions per second
r=10sec
(in 10 sec it comes to rest)
omega=0`
therefore `omega_0=-at`
`rarr a=-10/10=-`rev/s^2`
rarr omega=omega_0+at`
`=10-1xx5=5rev/s`
Therefore angular deceleration
`=1rev/s^2 and
Angular velocity after 5 sec is 5 rev/s.
Promotional Banner

Similar Questions

Explore conceptually related problems

A fan is rotating with a speed of 450 rec/minute. Afer being switched off it comes to rest in 10s. Assuming constant angular deceleration, calculate the number of recvolutions made by it before coming to rest.

To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration of a particle P (shown in figure) on the ring is (6hat(i)-8hat(j)) m//s^(2) . Find the angular acceleration of the ring & the abgular velocity at that instant. Radius of the ring is 2m.

A motor is rotating at a constant angular velocity of 600 rpm. The angular displacement in 2 second is …………..

A disc rotating about its axis, from rest it acquires a angular speed 100rev/s in 4 second. The angle rotated by it during these four seconds (in radian) is

A block of mass m = 20 kg is kept is a distance R = 1m from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, alpha = 3 rad//sec^(2) . The friction coefficient between block and table is mu = 0.5 . At time t = (x)/(30) from starting of motion (i.e. t =0) the block is just about to slip. Find the value of x (g = 10 m//s^(2))

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity omega is an example of non=inertial frame of reference. The relationship between the force vecF_(rot) experienced by a particle of mass m moving on the rotating disc and the force vecF_(in) experienced by the particle in an inertial frame of reference is vecF_(rot)=vecF_(i n)+2m(vecv_(rot)xxvec omega)+m(vec omegaxx vec r)xxvec omega . where vecv_(rot) is the velocity of the particle in the rotating frame of reference and vecr is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter fo a disc of radius R rotating counter-clockwise with a constant angular speed omega about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis (vecomega=omegahatk) . A small block of mass m is gently placed in the slot at vecr(R//2)hati at t=0 and is constrained to move only along the slot. The distance r of the block at time is

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity omega is an example of non=inertial frame of reference. The relationship between the force vecF_(rot) experienced by a particle of mass m moving on the rotating disc and the force vecF_(in) experienced by the particle in an inertial frame of reference is vecF_(rot)=vecF_(i n)+2m(vecv_(rot)xxvec omega)+m(vec omegaxx vec r)xxvec omega . where vecv_(rot) is the velocity of the particle in the rotating frame of reference and vecr is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter fo a disc of radius R rotating counter-clockwise with a constant angular speed omega about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis (vecomega=omegahatk) . A small block of mass m is gently placed in the slot at vecr(R//2)hati at t=0 and is constrained to move only along the slot. The distance r of the block at time is