Circular and Rotational Motion - 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:

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

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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 ball of mass is tied to a rope and moves along a horizontal circular path of radius as shown in the diagram (view from above). The maximum tension the rope can stand before breaking is given by $T_{max}$ . Which of the following represents the ball's linear velocity given that the rope does not break?

Answer

This is a centripetal force problem. In this case the tension on the rope is the centripetal force that keeps the ball moving on a circle.

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

If we want for the rope not to break, then the tension should never exceed $T_{max}$ .

$T_{max}\geq\frac{mv^2}{r}$

Now we just solve for velocity:

$v\leq\sqrt{\frac{rT_{max}}{m}}}$

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Question

A ball of mass is tied to a rope and moves along a horizontal circular path of radius as shown in the diagram (view from above). The maximum tension the rope can stand before breaking is given by $T_{max}$ . Which of the following represents the ball's linear velocity given that the rope does not break?

Answer

This is a centripetal force problem. In this case the tension on the rope is the centripetal force that keeps the ball moving on a circle.

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

If we want for the rope not to break, then the tension should never exceed $T_{max}$ .

$T_{max}\geq\frac{mv^2}{r}$

Now we just solve for velocity:

$v\leq\sqrt{\frac{rT_{max}}{m}}}$

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Question

A mechanic is using a wrench to loosen and tighten screws on an engine block and wants to increase the amount of torque he puts on the screws to adjust them more easily. Which of the following steps will not help him to do so?

Answer

Remember the torque equation:

$\tau = ||r|| * ||F|| * \sin \theta$

Exerting force on the wrench at an angle less perpendicular to the wrench will reduce $\sin \theta$ and thus reduce the torque.

The other answer options will all increase the torque being applied, making the twisting motion easier.

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Question

A mechanic is using a wrench to loosen and tighten screws on an engine block and wants to increase the amount of torque he puts on the screws to adjust them more easily. Which of the following steps will not help him to do so?

Answer

Remember the torque equation:

$\tau = ||r|| * ||F|| * \sin \theta$

Exerting force on the wrench at an angle less perpendicular to the wrench will reduce $\sin \theta$ and thus reduce the torque.

The other answer options will all increase the torque being applied, making the twisting motion easier.

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Question

Doing which of the following would allow you to find the center of mass of an object?

Answer

Center of mass can be found by spinning an object. It will naturally spin around its center of mass, due to the concept of even distribution of mass in relation to the center of mass. Shape and mass are important factors in this property, but the most improtant factor is the mass distribution.

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Question

If two masses, $m_{1}=50kg$ and $m_{2}=200kg$ are placed on a seesaw of length , where must the fulcrum be placed such that the seesaw remains level?

Answer

This question asks us to find the center of mass for this system. We know that the center of mass resides a distance $x_{1}$ from the first mass such that:

$m_{1}x_{1}=m_{2}(L-x_{1})$

In this case:

$m_{1}x_{1}+m_{2}x_{1}=m_{2}L$

$x_{1}(m_{1}+m_{2})=m_{2}L$

Plug in known values and solve.

$x_{1}=\frac{m_{2}L}{m_{1}+m_{2}}=\frac{200}{250}L=\frac{4}{5}L$

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Question

If the fulcrum of a balanced scale is shifted to the left, what type of adjustment must be made to rebalance the scale?

Answer

Changing the position of the fulcrum by moving it to the left means the center of mass will be to the right of the new position. Therefore, the scale will tip right. Adding more mass to the left end will rebalance the scale. None of the other options make sense. Adding more mass to the new fulcrum position will not change the balance of the scale because that mass is a negligible distance from the new fulcrum position and does nothing to change the masses on either side.

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Question

Three point masses are at the points , and

and a point mass is at the point .

How far from the origin is the center of mass of the system?

Answer

To find the center of mass, we have to take the weighted average of the x coordinates and the y coordinates.

Measures: Measures

First, we take the weighted measurement of the x-axis:

$x_c_m = \frac{(7.07\cdot40:g)+(-7.07 \cdot 40:g) + (0\cdot 40:g)+(0\cdot90:g)}{(3\cdot40)+90}$

We can see that the result of the x-axis contribution is equal to .

Now, let's look at the y-axis contribution:

$y_c_m = \frac{(0\cdot40:g)+(0 \cdot 40:g) + (-7.07\cdot 40:g)+(7.07\cdot90:g)}{(3\cdot40)+90}$

This equals to

Now that we have the x and y components, we take the root of squares to get the final answer:

$\uptext{center}= \sqrt{x^2+y^2}$

This will give us

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Question

Doing which of the following would allow you to find the center of mass of an object?

Answer

Center of mass can be found by spinning an object. It will naturally spin around its center of mass, due to the concept of even distribution of mass in relation to the center of mass. Shape and mass are important factors in this property, but the most improtant factor is the mass distribution.

Compare your answer with the correct one above

Question

If two masses, $m_{1}=50kg$ and $m_{2}=200kg$ are placed on a seesaw of length , where must the fulcrum be placed such that the seesaw remains level?

Answer

This question asks us to find the center of mass for this system. We know that the center of mass resides a distance $x_{1}$ from the first mass such that:

$m_{1}x_{1}=m_{2}(L-x_{1})$

In this case:

$m_{1}x_{1}+m_{2}x_{1}=m_{2}L$

$x_{1}(m_{1}+m_{2})=m_{2}L$

Plug in known values and solve.

$x_{1}=\frac{m_{2}L}{m_{1}+m_{2}}=\frac{200}{250}L=\frac{4}{5}L$

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Question

If the fulcrum of a balanced scale is shifted to the left, what type of adjustment must be made to rebalance the scale?

Answer

Changing the position of the fulcrum by moving it to the left means the center of mass will be to the right of the new position. Therefore, the scale will tip right. Adding more mass to the left end will rebalance the scale. None of the other options make sense. Adding more mass to the new fulcrum position will not change the balance of the scale because that mass is a negligible distance from the new fulcrum position and does nothing to change the masses on either side.

Compare your answer with the correct one above

Question

Three point masses are at the points , and

and a point mass is at the point .

How far from the origin is the center of mass of the system?

Answer

To find the center of mass, we have to take the weighted average of the x coordinates and the y coordinates.

Measures: Measures

First, we take the weighted measurement of the x-axis:

$x_c_m = \frac{(7.07\cdot40:g)+(-7.07 \cdot 40:g) + (0\cdot 40:g)+(0\cdot90:g)}{(3\cdot40)+90}$

We can see that the result of the x-axis contribution is equal to .

Now, let's look at the y-axis contribution:

$y_c_m = \frac{(0\cdot40:g)+(0 \cdot 40:g) + (-7.07\cdot 40:g)+(7.07\cdot90:g)}{(3\cdot40)+90}$

This equals to

Now that we have the x and y components, we take the root of squares to get the final answer:

$\uptext{center}= \sqrt{x^2+y^2}$

This will give us

Compare your answer with the correct one above

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