A particular grinding wheel is a uniform solid disk of radius 8cm and mass 3kg. While the wheel is spinning at 50rad/s. A tool is pressed against its rim in a way that causes a friction force of 40N. If the disk is just "coasting" when the torque is applied, at what rate will its angular velocity start to decrease?
To find the rate at which the angular velocity of the grinding wheel will start to decrease, we need to calculate the net torque acting on the wheel.
The net torque τ is equal to the product of the friction force F and the radius r at which the force is applied:
τ = F * r
Given:
Friction force, F = 40 N
Radius, r = 8 cm = 0.08 m
Plugging in the values, we have:
τ = 40 N * 0.08 m
τ = 3.2 Nm
Now, the torque τ is also equal to the product of the moment of inertia I and the angular acceleration α:
τ = I * α
The moment of inertia for a solid disk rotating about its axis is given by the formula:
I = 0.5 * m * r^2
Given:
Mass, m = 3 kg
Radius, r = 0.08 m
Plugging in the values, we have:
I = 0.5 * 3 kg * (0.08 m)^2
I = 0.012 kgm^2
Substituting the moment of inertia I in the torque equation, we get:
τ = 0.012 kgm^2 * α
Now, rearranging the equation, we can solve for the angular acceleration α:
α = τ / I
α = (3.2 Nm) / (0.012 kgm^2)
α = 266.67 rad/s^2
The angular acceleration α represents the rate at which the angular velocity will start to decrease.
To find the rate at which the angular velocity of the grinding wheel will decrease, we need to consider the torque acting on the wheel and apply Newton's second law for rotational motion.
The torque on the wheel is caused by the friction force applied at its rim. Torque is given by the equation:
τ = r * F * sin(θ),
where:
τ is the torque,
r is the radius at which the force is applied,
F is the force being applied, and
θ is the angle between the force vector and the lever arm.
In this case, the force is 40N and the radius of the wheel is 8cm (0.08m). Since the force is applied tangentially to the rim, the angle θ will be 90 degrees (sin(90 degrees) = 1). Plugging in the values, we get:
τ = (0.08m) * (40N) * 1 = 3.2 Nm.
The torque acting on the wheel is 3.2 Nm.
Next, we apply Newton's second law for rotational motion:
τ = I * α,
where:
τ is the torque,
I is the moment of inertia of the wheel, and
α is the angular acceleration.
The moment of inertia of a solid disk can be calculated using the equation:
I = (1/2) * m * r^2,
where:
m is the mass of the wheel, and
r is the radius of the wheel.
In this case, the mass of the wheel is 3kg and the radius is 8cm (0.08m). Plugging in the values, we get:
I = (1/2) * (3kg) * (0.08m)^2 = 0.0096 kgm^2.
The moment of inertia of the wheel is 0.0096 kgm^2.
Now we can solve for the angular acceleration (α):
3.2 Nm = (0.0096 kgm^2) * α.
Rearranging the equation, we find:
α = (3.2 Nm) / (0.0096 kgm^2) ≈ 333.33 rad/s^2.
The angular acceleration of the wheel is approximately 333.33 rad/s^2.
Finally, we know that the rate of change of angular velocity (angular deceleration) is equal to the angular acceleration (α). Therefore, the angular velocity will start to decrease at a rate of approximately 333.33 rad/s^2.