Circular Motion Concepts and Equations - AP Physics C Electricity & Magnetism

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Question

A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?

Answer

The ball is experiencing centripetal force so that it can travel in a circular path. This centripetal force is written as the equation below.

$F_c=ma=\frac{mv^2}{r}$

Remember that centripetal acceleration is given by the following equation.

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

Since the centripetal force is coming from the tension of the string, set the tension force equal to the centripetal force.

$T=\frac{mv^2}{r}$

Since we're trying to find the speed of the ball, we solve for v.

$v=\sqrt{\frac{Tr}{m}}$

We know the following information from the question.

We can use this information in our equation to solve for the speed of the ball.

$v=\sqrt{\frac{(87N)(1.8m)}{1.6kg}}}$

$v=9.9\frac{m}{s}$

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Question

In uniform circular motion, the net force is always directed ___________.

Answer

The correct answer is "toward the center of the circle." Newton's second law tells us that the direction of the net force will be the same as the direction of the acceleration of the object.

In uniform circular motion, the object accelerates towards the center of the circle (centripetal acceleration); the net force acts in the same direction.

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Question

A car moves around a circular path of radius 100m at a velocity of $20\frac{m}{s}$ . What is the coefficient of friction between the car and the road?

Answer

The force of friction is what keeps the car in circular motion, preventing it from flying off the track. In other words, the frictional force will be equal to the centripetal force.

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

$F_f=\mu F_N=\mu mg$

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

We can cancel mass from either side of the equation and rearrange to solve for the coefficient of friction:

$\frac{v^2}{r}=g\mu$

$\mu=\frac{v^2}{rg}$

We can use our given values to solve:

$\mu=\frac{(20\frac{m}{s})^2}{(100m)(9.8\frac{m}{s^2})}=0.4$

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Question

An object of mass 10kg undergoes uniform circular motion with a constant velocity of $10\frac{m}{s}$ at a radius of 10m. How long does it take for the object to make one full revolution?

Answer

The time for one full revolution can be calculated simply by manipulating the defintion of velocity, where the distance is just the circumference of the circlular path. The time it takes is modeled by the following equation:

$t=\frac{d}{v}=\frac{2\pi r}{v}$

Use the given radius and velocity to solve for the time per revolution:

$t=\frac{2\pi(10m)}{10\frac{m}{s}}=2\pi\approx6.3s$

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Question

A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?

Answer

The ball is experiencing centripetal force so that it can travel in a circular path. This centripetal force is written as the equation below.

$F_c=ma=\frac{mv^2}{r}$

Remember that centripetal acceleration is given by the following equation.

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

Since the centripetal force is coming from the tension of the string, set the tension force equal to the centripetal force.

$T=\frac{mv^2}{r}$

Since we're trying to find the speed of the ball, we solve for v.

$v=\sqrt{\frac{Tr}{m}}$

We know the following information from the question.

We can use this information in our equation to solve for the speed of the ball.

$v=\sqrt{\frac{(87N)(1.8m)}{1.6kg}}}$

$v=9.9\frac{m}{s}$

Compare your answer with the correct one above

Question

In uniform circular motion, the net force is always directed ___________.

Answer

The correct answer is "toward the center of the circle." Newton's second law tells us that the direction of the net force will be the same as the direction of the acceleration of the object.

In uniform circular motion, the object accelerates towards the center of the circle (centripetal acceleration); the net force acts in the same direction.

Compare your answer with the correct one above

Question

A car moves around a circular path of radius 100m at a velocity of $20\frac{m}{s}$ . What is the coefficient of friction between the car and the road?

Answer

The force of friction is what keeps the car in circular motion, preventing it from flying off the track. In other words, the frictional force will be equal to the centripetal force.

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

$F_f=\mu F_N=\mu mg$

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

We can cancel mass from either side of the equation and rearrange to solve for the coefficient of friction:

$\frac{v^2}{r}=g\mu$

$\mu=\frac{v^2}{rg}$

We can use our given values to solve:

$\mu=\frac{(20\frac{m}{s})^2}{(100m)(9.8\frac{m}{s^2})}=0.4$

Compare your answer with the correct one above

Question

An object of mass 10kg undergoes uniform circular motion with a constant velocity of $10\frac{m}{s}$ at a radius of 10m. How long does it take for the object to make one full revolution?

Answer

The time for one full revolution can be calculated simply by manipulating the defintion of velocity, where the distance is just the circumference of the circlular path. The time it takes is modeled by the following equation:

$t=\frac{d}{v}=\frac{2\pi r}{v}$

Use the given radius and velocity to solve for the time per revolution:

$t=\frac{2\pi(10m)}{10\frac{m}{s}}=2\pi\approx6.3s$

Compare your answer with the correct one above

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