A small 520-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1.5 m. Calculate the moment of inertia of the ball about the center of the circle. Then calculate the torque needed to keep the ball rotating at constant angular velocity if air resistance exerts a force of 0.015 N on the ball. Ignore the rod's moment of inertia and air resistance. I don't understand how to do this problem, please help with step-by-step.

Why did the ball bring a calculator to the party? Because it wanted to calculate the moment of inertia and torque, of course! Let's get into it.

Step 1: Calculate the moment of inertia of the ball about the center of the circle.
Moment of inertia (I) represents an object's resistance to changes in rotational motion. For a point mass rotating about an axis that passes through its center of mass, the moment of inertia is given by:

I = m * r^2

Where:
m = mass of the ball (in kilograms)
r = radius of the circle (in meters)

In this case, we are given the mass of the ball as 520 grams (which we need to convert to kilograms), and the radius of the circle is 1.5 meters. So let's convert grams to kilograms and calculate the moment of inertia:

Convert 520 grams to kilograms:
520 grams = 520/1000 = 0.52 kg

Calculate the moment of inertia:
I = (0.52 kg) * (1.5 m)^2
= 0.52 * 2.25
= 1.17 kg·m^2

So the moment of inertia of the ball about the center of the circle is 1.17 kg·m^2.

Step 2: Calculate the torque needed to keep the ball rotating at constant angular velocity.
Torque (τ) is the rotational equivalent of force and is given by:

τ = r * F

Where:
r = radius of the circle (in meters)
F = force applied perpendicular to the line of action to create rotation (in Newtons)

In this case, we are given that air resistance exerts a force of 0.015 N on the ball. Let's calculate the torque:

τ = (1.5 m) * (0.015 N)
= 0.0225 N·m

So, the torque needed to keep the ball rotating at a constant angular velocity, despite the air resistance, is 0.0225 N·m.

Keep spinning those questions!

To solve this problem, we need to use the formula for moment of inertia and torque.

Step 1: Calculate the moment of inertia of the ball about the center of the circle.
The moment of inertia of a point mass rotating about an axis perpendicular to its mass is given by the formula:
I = m * r^2
where I is the moment of inertia, m is the mass of the object, and r is the radius of the rotation.

In this case, the mass of the ball is 520 grams, which is equivalent to 0.520 kg. The radius of the rotation is 1.5 m. Substitute these values into the formula to calculate the moment of inertia:
I = 0.520 kg * (1.5 m)^2
I = 0.520 kg * 2.25 m^2
I = 1.17 kg * m^2

Step 2: Calculate the torque needed to keep the ball rotating at constant angular velocity.
The torque needed to keep an object rotating at a constant angular velocity is equal to the product of the angular acceleration of the object and its moment of inertia.

In this case, we need to calculate the torque exerted by air resistance to counteract the torque due to the ball's weight.

The force exerted by air resistance is given as 0.015 N. Since the ball is rotating in a horizontal circle, the torque due to air resistance is equal to this force multiplied by the radius of the circle:
τ = F * r
where τ is the torque, F is the force, and r is the radius.

Substitute the values into the formula to calculate the torque:
τ = 0.015 N * 1.5 m
τ = 0.0225 Nm

Since the torque needed to keep the ball rotating at a constant angular velocity is equal to the torque due to air resistance, the torque is 0.0225 Nm.

Therefore, the moment of inertia of the ball about the center of the circle is 1.17 kg * m^2, and the torque needed to keep the ball rotating at constant angular velocity is 0.0225 Nm.

To calculate the moment of inertia of the ball about the center of the circle, you can use the formula:

I = mr^2

Where:
I is the moment of inertia,
m is the mass of the ball, and
r is the radius of the circle.

1. Determine the values:
- Mass of the ball (m) = 520 grams = 0.520 kg
- Radius of the circle (r) = 1.5 m

2. Plug in the values into the formula:
I = 0.520 kg * (1.5 m)^2

3. Calculate:
I = 0.520 kg * 2.25 m^2
I = 1.170 kg m^2

So, the moment of inertia of the ball about the center of the circle is 1.170 kg m^2.

To calculate the torque needed to maintain the angular velocity, you can use the formula:

τ = I * α

Where:
τ is the torque,
I is the moment of inertia, and
α is the angular acceleration.

In this case, the angular velocity is constant, so the angular acceleration (α) is zero. Therefore, the torque needed is also zero. The air resistance force (0.015 N) does not contribute to the torque because it acts in the opposite direction of rotation.

Hence, the torque needed to keep the ball rotating at a constant angular velocity is zero.

The moment of inertia for the mass is

I=mr^2

Now, torque to counter air friction..

airresiatance*r=torque