Question

A constant torque of 200Nm turns a wheel about its centre. The M.I/ about the axis is 100 kgm^2. Find the angular velocity gained in 4s and K.E gained after 20 revolution

Answer 1

1 answer · Physics

Best Answer

1. A constant torque of 200N m turns a wheel about its centre. The moment of inertia about this axis is 100kg m^2. Find

(a) the angular velocity gained in 4s

Torque = inertia x alpha (alpha is angular acceleration, rad/s^2)

200 N.m / 100 kg.m^2 = 2 rad/s^2

angular speed = omega (rad/s) = alpha x time = 2 x 4 = 8 rad/s (answer)

(b) the kinetic energy gained after 20 revs

work done by torque = torque x distance in radians.

20 revs x 2pi = 125.66 radians

torque = 200 N.m

200 x 125.66 = 25132.74 J (answer)

2. A flywheel has a kinetic energy of 200J. Calculate the number of revolutions it makes before coming to rest if a constant opposing couple of 5N m is applied to the flywheel.

as above, use work done by torque, but in this case it will be work done by frictional torque, the 5 N.m should actually be -5 N.m, but still, i'll stick to the question.

200/5 = 40 radians

40/(2pi) = 6.366 revolutions (answer)

3. If the moment of inertia of the flywheel about its centre is 4kg m^2, how long does it take to come to rest.

Rotational KE = 200 J

RKE = 1/2Iω^2

200/(0.5 x 4) = ω^2 = 100, sq-root(100) = ω = 10 rad/s

40 radians / ((10 + 0)/2) = time = 8 secs

Answer 2

Answer:

(a). The angular velocity is 8 rad/s.

(b). The kinetic energy gained after 20 revolution is 25132.74 J.

Explanation:

Given that,

Torque = 200 Nm

Moment of inertia = 100 kgm^2

Time = 4 sec

We need to calculate the angular acceleration

Using formula of torque

$\tau=I\times\alpha$

$\alpha=\dfrac{\tau}{I}$

Put the value into the formula

$\alpha=\dfrac{200}{100}$

$\alpha=2\ rad/s^2$

We need to calculate the angular velocity

Using formula of angular velocity

$\omega=\alpha\times t$

Put the value into the formula

$\omega=2\times4$

$\omega=8\ rad/s$

(b). We need to calculate the kinetic energy gained after 20 revolution

Using formula of the kinetic energy

$Work\ done\ by\ torque=torque\times distance$

$W=\tau\times\omega$

Put the value into the formula

$W=200\times20\times2\pi$

$W=25132.74\ J$

Hence, (a). The angular velocity is 8 rad/s.

(b). The kinetic energy gained after 20 revolution is 25132.74 J.