# A cylindrical centrifuge of mass 8 kg and radius 10 cm spins at a speed of 80,000 rpm. Calculate the minimum braking torque that must be applied to stop the rotor within 30 s from the instant the motor is turned off.

Hello!

I suppose we apply a constant braking force `F` perpendicularly to the radius of spin. Then the torque `tau` (moment of force) will be `F*r.`

From the other hand, this torque will cause an angular acceleration (deceleration in this case) `alpha,` and the relation between them is

`tau=I*alpha,` where `I` is the moment of inertia of a centrifuge.

For an object whose mass is at the constant distance from the axis of rotation, the moment of inertia is `I=m*r^2.`  Linking this together, obtain

`F*r=m*r^2*alpha,` or `F=m*r*alpha.`

The angular speed will decrease uniformly at a rate `alpha,` `V=V_0-alpha t.` A centrifuge stops when `V=0,` so `alpha=V_0/t_1,` where `V_0` is the initial angular speed and `t_1` is the given time of braking.

Finally, `F=m*r*V_0/t_1,` all these quantities are given. The only problem is that `V_0` is given in rpm, while it is required in radians per second. One rpm is `(2pi)/60` radians per second, so the final answer is

`8*0.1*80000*(2pi)/60*1/30=8*80*pi/9 approx 223(N).`

This is the braking force. The torque itself is `F*r approx 22.3 (N*m).`

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