A pottery wheel with rotational inertia 40 kgm^2 is rotating at 10 rev/s. A 4 kg

Question

A pottery wheel with rotational inertia 40 kgm^2 is rotating at 10 rev/s. A 4 kg lump of clay is thrown straight down onto the wheel 1.2 m from the axis of rotation. Whats the angular momentum? Whats the angular speed after the clay lands? How much torque must be applied if you wish to bring it to a stop in 3s? A pottery wheel with rotational inertia 40 kgm^2 is rotating at 10 rev/s. A 4 kg lump of clay is thrown straight down onto the wheel 1.2 m from the axis of rotation. Whats the angular momentum? Whats the angular speed after the clay lands? How much torque must be applied if you wish to bring it to a stop in 3s?

Explanation / Answer

here,

moment of inertia , I = 40 kg.m^2

initial angular speed , wi = 10 rev/s = 62.8 rad/s

the angular momentum , Li = I * wi

Li = 40 * 62.8 kg.m^2.rad/s

Li = 2512 kg.m^2.rad/s

let the final angular speed be wf

using conservation of angular momentum

Li = ( I + 4 * 1.2^2) * wf

2512 = ( 40 + 4 * 1.2^2) * wf

wf = 54.9 rad/s

let the angular accelration for stopping in time (t =3s ) be alpha

0 = wf + alpha * t

0 = 54.9 + alpha * 3

alpha = - 18.3 rad/s

the torque must be applied if you wish to bring it to a stop , T = (I + 4 * 1.2^2) * alpha

T = ( 40 + 4 * 1.2^2) * (-18.3) N.m

T = - 837.4 N.m

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