A 715 gram grinding wheel 29.5 {\rm cm} in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 210 {\rm rpm} about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 46.5 {\rm s} with constant angular acceleration due to friction at the axle.

Question

A 715 gram grinding wheel 29.5 {\rm cm} in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 210 {\rm rpm} about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 46.5 {\rm s} with constant angular acceleration due to friction at the axle.

Answer 1

and the question ?

Answer 2

To find the net torque acting on the grinding wheel when it stops, we can use the formula:

Net Torque = Moment of Inertia * Angular Acceleration

First, let's find the moment of inertia of the grinding wheel. The moment of inertia for a uniform disk rotating about an axis perpendicular to its face is given by the formula:

Moment of Inertia = (1/2) * mass * radius^2

Given:
Mass of the grinding wheel (m) = 715 grams = 0.715 kg
Radius of the grinding wheel (r) = 29.5 cm = 0.295 m

Plugging these values into the moment of inertia formula:

Moment of Inertia = (1/2) * 0.715 kg * (0.295 m)^2

Simplifying:

Moment of Inertia = 0.5 * 0.715 kg * 0.086825 m^2
Moment of Inertia = 0.0293197 kg*m^2 (approximately)

Next, let's find the angular acceleration of the grinding wheel. We can use the kinematic equation for rotational motion:

θ = ω_0*t + (1/2) * α * t^2

where:
θ - Angle rotated (in radians)
ω_0 - Initial angular velocity
α - Angular acceleration
t - Time

In this case, the angle rotated (θ) is given by 2π radians (one complete revolution). The initial angular velocity (ω_0) is 210 rpm (converted to radians per second).

θ = 2π radians
ω_0 = 210 rpm * (2π/60) rad/s = 21.99 rad/s

Plugging in these values, we can solve for α:

2π = 21.99 rad/s * 46.5 s + (1/2) * α * (46.5 s)^2
2π = 21.99 rad/s * 46.5 s + 10.8225 s^2 * α

Rearranging the equation:

20π.5 rad/s * 46.5 s = 10.8225 s^2 * α

Simplifying:

α = (20π.5 rad/s * 46.5 s) / 10.8225 s^2
α = 39.611 rad/s^2 (approximately)

Now, plug the values of the moment of inertia (I) and the angular acceleration (α) into the equation to find the net torque (τ):

Net Torque = Moment of Inertia * Angular Acceleration
τ = 0.0293197 kg*m^2 * 39.611 rad/s^2

Calculating the net torque:

τ = 1.16 N*m (approximately)

Therefore, the net torque acting on the grinding wheel when it stops is approximately 1.16 N*m.