Question

The figure below shows, at left, a solid disk of radius R 0.700 m and mass 75.0 kg. Mounted directly to it and coaxial with it is a pulley with a much smaller mass and a radius ofr 0.230 m. The disk and pulley assembly are on a frictionless axle. A belt is wrapped around the pulley and connected to an electric motor as shown on the right. The turning motor gives the disk and pulley a clockwise angular acceleration of 1.67 rad/s2. The tension Tu in the upper (taut) segment of the belt is 165 N (a) What is the tension (in N) in the lower (slack) segment of the belt? (b) What If? You replace the belt with a different one (one slightly longer and looser, but still tight enough that it does not sag). You again turn on the motor so that the disk accelerates clockwise. The upper segment of the belt once again has a tension of 165 N, but now the tension in the lower belt is exactly zero. What is the magnitude of the angular acceleration (in rad/s2)? ) rad/s2

Answer 1

The tension in the lower segment of the belt is approximately 577.91 N. The magnitude of the angular acceleration is zero in the second scenario when the tension in the lower belt is zero.

Step-by-step explanation:

(a) Tension in the lower segment of the belt:

To find the tension in the lower segment of the belt, we can use the equation for the net torque.

In this case, the net torque is given by the product of the moment of inertia and the angular acceleration:

Torque = Moment of Inertia * Angular Acceleration

The moment of inertia of the system can be calculated by adding the moments of inertia of the disk and the pulley:

Moment of Inertia = Moment of Inertia of Disk + Moment of Inertia of Pulley

Using the formulas for the moments of inertia of a disk and a pulley:

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

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

Substituting the given values:

• Mass of disk = 75.0 kg
• Mass of pulley = much smaller mass, so we can neglect it
• Radius of disk = 0.700 m
• Radius of pulley = 0.230 m
• Angular acceleration = 1.67 rad/s^2

Plugging in these values, we get:

Moment of Inertia = (1/2) * 75.0 kg * (0.700 m)^2 + (1/2) * much smaller mass * (0.230 m)^2

Since the mass of the pulley is much smaller, we can neglect it and simplify the equation to:

Moment of Inertia = (1/2) * 75.0 kg * (0.700 m)^2

Next, we can find the torque:

Torque = Moment of Inertia * Angular Acceleration

Plugging in the values for the moment of inertia and angular acceleration, we get:

Torque = (1/2) * 75.0 kg * (0.700 m)^2 * 1.67 rad/s^2

Simplifying this expression, we find that the torque is approximately 132.72 Nm.

Now, we can find the tension in the lower segment of the belt using the equation for torque:

Torque = Tension * Radius

Plugging in the values for the torque and radius, we get:

132.72 Nm = Tension * 0.230 m

Solving for Tension, we find that the tension in the lower segment of the belt is approximately 577.91 N.

(b) Magnitude of the angular acceleration:

In the second scenario, when the tension in the lower belt is zero, we can use the same equations for torque and moment of inertia to find the angular acceleration.

Setting the torque equal to zero, we get:

0 = Tension * 0.230 m

Since the tension is zero, the torque and angular acceleration are also zero.

Therefore, the magnitude of the angular acceleration is zero.