Final answer:
The torque exerted by friction on the grinding wheel while it is slowing down is approximately -0.0060 N*m.
Step-by-step explanation:
To calculate the torque exerted by friction while the grinding wheel is slowing down, we need to use the equations of rotational motion.
From the given information, we know that the mass of the grinding wheel is 750 grams (0.75 kg) and its diameter is 25.0 cm (0.25 m).
The moment of inertia for a solid disk can be calculated using the equation:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass, and r is the radius.
Plugging in the given values:
I = (1/2) * 0.75 kg * (0.25 m/2)^2 = 0.01171875 kg*m^2
The angular acceleration can be calculated using the equation:
α = (ωf - ωi) / t
where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity (which is given as 220 rpm), and t is the time taken.
Plugging in the given values:
ωf = 0 rad/s (since the wheel stops), ωi = 220 rpm * (2π rad/rev) / 60 s = 23.094 rad/s, and t = 45.0 s
α = (0 - 23.094 rad/s) / 45.0 s = -0.5132 rad/s^2
The torque exerted by friction can be calculated using the equation:
τ = I * α
Plugging in the calculated values:
τ = 0.01171875 kg*m^2 * -0.5132 rad/s^2
τ ≈ -0.0060 N*m
Therefore, the torque exerted by friction while the grinding wheel is slowing down is approximately