Physics - Using Circular Motion Equations

A baseball has a radius of . What is the moment of inertia for the ball?

$I_{sphere}=\frac{2}{5}mr^2$

$8.7*10^{-4}kg*m^2$

$2.3*10^{-2}kg*m^2$

$5.7*10^{-2}kg*m^2$

$4.3*10^{-4}kg*m^2$

Explanation

The given equation for moment of inertia is:

$I_{sphere}=\frac{2}{5}mr^2$

Use the given values for the mass and radius of the ball to solve for the moment of inertia.

$I=\frac{2}{5}mr^2$

$I=\frac{2}{5}(1.5kg)(0.038m)^2$

$I=8.7*10^{-4}kg*m^2$

A car makes a right turn. The radius of this curve is . If the force of friction between the tires and the road is , what is the maximum velocity that the car can have before skidding?

$3.01\frac{m}{s}$

$9.06\frac{m}{s}$

$0.81\frac{m}{s}$

$2.16\frac{m}{s}$

$82\frac{m}{s}$

Explanation

To solve this problem, recognize that the force due to friction must equal the centripetal force of the curve:

This will give the maximum force that the car can have in the curve without skidding. Expand the equation for centripetal force.

$F_f=m\frac{v^2}{r}$

We are given the value for the force of friction, the mass of the car, and the radius of the curve. Using these values, we can find the velocity.

$2587.2N=1200kg*\frac{v^2}{4.2m}$

$\frac{2587.2N}{1200kg}=\frac{v^2}{4.2m}$

$2.156\frac{m}{s^2}=\frac{v^2}{4.2m}$

$4.2m*2.156\frac{m}{s^2}=v^2$

$9.0552\frac{m^2}{s^2}=v^2$

$\sqrt{9.0552\frac{m^2}{s^2}}=\sqrt{v^2}$

$3.01\frac{m}{s}=v$

How much force is required for a hammer to produce of torque?

Explanation

The formula for torque is:

$\tau=F*r$

We are given the length of the hammer (radius of the swing) and the torque produced. Using these values, we can solve for the force required.

$\frac{4.2N*m}{0.3m}=F$

A child swings a pail of water in a circle, using her shoulder as the pivot point. If the child's arm is long and the pail of water has a linear momentum of $1.67\frac{kg\cdot m}{s}$ , what is the angular momentum of the pail of water?

$1.3\ \frac{kg\cdot m^2}{s}$

$2.1\ \frac{kg\cdot m^2}{s}$

$0.87\ \frac{kg\cdot m^2}{s}$

$1.07\ \frac{kg\cdot m^2}{s}$

$2.23\ \frac{kg\cdot m^2}{s}$

Explanation

Angular momentum is equal to the linear momentum times the radial arm.

Since the pivot point of the child is the shoulder, the radial arm is the child's arm length. We are given this length, as well as the linear momentum, allowing us to solve for the angular momentum.

$L=(0.8m)(1.67\frac{kg\cdot m}{s})$

$L=1.3\frac{kg\cdot m^2}{s}$

A rock is swung in a circle on a long rope. How much centripetal force is required for the rock to maintain a velocity of $0.9\frac{m}{s}$ ?

Explanation

The equation for centripetal force is:

$F_c=m\frac{v^2}{r}$

We are given the radius (length of the rope), velocity, and mass of the rock, allowing us to calculate the centripetal force.

$F_c=0.22kg*\frac{(0.9\frac{m}{s})^2}{0.8m}$

$F_c=0.22kg*\frac{0.81\frac{m^2}{s^2}}{0.8m}$

$F_c=0.22kg*1.0125\frac{m}{s^2}$

A object is moving in a perfect circle with a radius of and has a linear momentum of $35\frac{kg*m}{s}$ .

What is the centripetal acceleration on the object?

$81.65\frac{m}{s^2}$

$253.13\frac{m}{s^2}$

$5.13\frac{m}{s^2}$

$49.32\frac{m}{s^2}$

$26.34\frac{m}{s^2}$

Explanation

Centripetal acceleration is equal to the tangential velocity squared over the radius:

$a_c=\frac{v^2}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

$35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$

Use the linear velocity and the radius in the initial equation to solve for the centripetal acceleration.

$a_c=\frac{v^2}{r}$

$a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$

$a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$

$a_c=81.65\frac{m}{s^2}$

A object is moving in a perfect circle with a radius of and has a linear momentum of $35\frac{kg*m}{s}$ .

What is the moment of inertia on this object?

Explanation

Since the object is moving in a perfect circle and not rotating about some fixed point within itself (like spinning a ball or a frisbee), the equation for moment of inertia is:

We are given the mass and radius, allowing us to calculate the moment of inertia from this equation.

A object is moving in a perfect circle with a radius of and has a linear momentum of $35\frac{kg*m}{s}$ .

What is the centripetal force on the object?

Explanation

Newton's second law states that . We know the mass of the object, but we need to find the centripetal acceleration to calculate the centripetal force.

Centripetal acceleration is equal to the tangential velocity squared over the radius:

$a_c=\frac{v^2}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

$35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$

Use the linear velocity and the radius in the previous equation to solve for the centripetal acceleration.

$a_c=\frac{v^2}{r}$

$a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$

$a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$

$a_c=81.65\frac{m}{s^2}$

Use the centripetal acceleration in Newton's second law, along with the mass, to calculate the centripetal force.

$F_c=2.2kg*81.65\frac{m}{s^2}$

An object moves in a circle with a constant velocity of $3\frac{m}{s}$ . If the radius of the circle is , what is the centripetal acceleration on the object?

$4.5\frac{m}{s^2}$

$9\frac{m}{s^2}$

$18\frac{m}{s^2}$

$1.5\frac{m}{s^2}$

$6\frac{m}{s^2}$

Explanation

The formula for centripetal acceleration is:

$a_c=\frac{v^2}{r}$

We are given the velocity and the radius, allowing us to solve for the acceleration.

$a_c=\frac{v^2}{r}$

$a_c=\frac{(3\frac{m}{s})^2}{2m}$

$a_c=\frac{9\frac{m^2}{s^2}}{2m}$

$a_c=4.5\frac{m}{s^2}$

A top spins so that the angular velocity is $\small 31\frac{rad}{s}$ . If the top has a radius of , what is the centripetal acceleration acting on a point on the edge of the top?