An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel. The flywheel is a solid disk with a mass of 80.0kg and a dimaeter of 1.25m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230m. If the tension in the upper (taut) segment of the belt is 135 N and the flywheel has a clockwise angular acceleration of 1.67 rad/s^2, find the tension in the lower (slack) segment of the belt.

Sol: This problem seems to involve torque; I think solid disc Icm (moment of intertia) is 1/2MR^2; there are other unknown I'm sure but if someone can walk me through this problem (step by step), I'd be most appreciative. I have a test tomorrow. I tried and tried, but finally gave up as I must move on to other practice problems.

force net= upper force - lower force

But force net* radiuspulley= Itotal*angacceleration.

Now I total is the sum of the flywheel and the pulley.

Solve for lower force.

Itotal = Ipulley + Iflywheel

Ipulley = mr^2

Iflywheel = 1/2 mr^2

Itotal = mr^2 + 1/2 mr^2

Itotal = 1.5 mr^2

Force net * radiuspulley = 1.5 mr^2 * angacceleration

Force net * radiuspulley = 1.5 * 80 * 0.230^2 * 1.67

Force net * radiuspulley = 58.904

Force net = 58.904 / 0.230

Force net = 256.3 N

Lower force = Upper force - Force net

Lower force = 135 - 256.3

Lower force = -121.3 N

To solve this problem step-by-step, we'll start by finding the moment of inertia (I) of the flywheel and the pulley.

1. Flywheel moment of inertia (I_flywheel):

Since the flywheel is a solid disk, we can use the formula for the moment of inertia of a solid disk about its central axis, which is given by:

I = (1/2) * mass * radius^2

Substituting the given values, we have:

I_flywheel = (1/2) * 80.0 kg * (0.625 m)^2
= 15.625 kg·m^2

2. Pulley moment of inertia (I_pulley):

Since the pulley has a smaller mass and radius, we can approximate its moment of inertia as negligible compared to the flywheel's moment of inertia.

Next, let's find the torque (T) exerted on the flywheel:

T = I * angular acceleration
= (15.625 kg·m^2) * (1.67 rad/s^2)
= 26.014 N·m

Now, we can apply Newton's second law for rotational motion, which states that the net torque acting on an object is equal to the moment of inertia times the angular acceleration:

T = (upper force - lower force) * radius_pulley

Rearranging the equation to solve for the lower force (lower_force):

lower_force = (T / radius_pulley) - upper force

Plugging in the given values:

lower_force = (26.014 N·m / 0.230 m) - 135 N
= 113.139 N - 135 N
= -21.861 N

Note: The negative sign indicates that the lower force is in the opposite direction of the upper force. Since the problem statement asks for the tension, we take the magnitude of the lower force as:

|lower_force| = 21.861 N

Therefore, the tension in the lower (slack) segment of the belt is approximately 21.861 N.

To solve this problem, we can start by considering the torque equation that relates the torque exerted on an object to its moment of inertia and angular acceleration:

Torque = Moment of Inertia * Angular Acceleration

In our case, the torque exerted on the flywheel is given by:

Torque = Tension in the upper segment of the belt * Radius of the pulley attached to the flywheel

We can rewrite this equation as:

Tension in the upper segment of the belt = (Torque / Radius of the pulley attached to the flywheel)

Now, the moment of inertia of the flywheel can be calculated using the formula for a solid disk:

Moment of Inertia (I) = (1/2) * Mass * Radius^2

Substituting the given values, we have:

I = (1/2) * 80.0 kg * (1.25 m/2)^2

Next, we need to find the moment of inertia of the pulley. Since it has a much smaller mass compared to the flywheel, we can approximate its moment of inertia as zero.

Now, we know that the total moment of inertia (Itotal) is the sum of the moment of inertia of the flywheel and the pulley. In this case, Itotal = I.

So, the torque exerted on the flywheel can be written as:

Torque = Tension in the upper segment of the belt * Radius of the pulley attached to the flywheel

Using the torque equation, we have:

Tension in the upper segment of the belt * Radius of the pulley attached to the flywheel = Itotal * Angular Acceleration

Now, we can solve for the tension in the lower segment of the belt. Rearranging the equation, we have:

Tension in the lower segment of the belt = (Itotal * Angular Acceleration) / Radius of the pulley attached to the flywheel

Substituting the known values, we have:

Tension in the lower segment of the belt = (I * Angular Acceleration) / Radius of the pulley attached to the flywheel

Calculating these values will give you the tension in the lower segment of the belt.