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AP Physics 1 : Circular and Rotational Motion

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Example Question #51 : Circular And Rotational Motion

A 250 g ball is being swung faster and faster in a horizontal circle. When the speed reaches $15 \frac{m}{s}$ , the tension () is 500 N and the string breaks. What is the radius of the circle?

Possible Answers:

5.25 m

11.25 cm

1.7 m

13.3 cm

0.75 cm

Correct answer:

11.25 cm

Explanation:

In order to find the radius of the circle that corresponds to the maximum tension the string can experience without breaking, we must rearrange the centripetal force equation for objects moving in a circle:

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

$F_c = T_{max}\Rightarrow r= m \frac{v^2}{T_{max}}$ $=.25kg\left ( \frac{(15\frac{m}{s})^2}{500N} \right ) = 11.25cm$

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Example Question #51 : Circular And Rotational Motion

You are riding a rollercoaster going around a vertical loop, on the inside of the loop. If the loop has a radius of 50 m , how fast must the cart be moving in order for you to feel three times as heavy at the top of the loop?

Possible Answers:

$31.3 \frac{m}{s}$

$38.4 \frac{m}{s}$

$44.3 \frac{m}{s}$

$49.5 \frac{m}{s}$

$6.70 \frac{m}{s}$

Correct answer:

$44.3 \frac{m}{s}$

Explanation:

To solve this problem, it is important to remember the signs of each force acting on the coaster. Since we are inside the loop at the top, both the weight and the normal force point toward to center of the circular path, and are thus positive.

$mg + N = m \frac{v^2}{r}$

Next, we note that the normal force is what a rider "feels" during their ride on the rollercoaster. We want the speed at which a rider would feel three times as heavy, or three times their weight.

$mg+N=mg+3mg=m\frac{v^2}{r}\Rightarrow v=\sqrt{4gr}=\sqrt{4(9.81\frac{m}{s^2})50m}=44.3\frac{m}{s}$

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Example Question #53 : Circular And Rotational Motion

A car's tire can rotate at a frequency of 3000rpm (revolutions per minute). Given that a typical tire radius is .5m , what is the centripetal acceleration of the tire?

Possible Answers:

$5000\pi^2\frac{m}{s^2}$

$2500\pi^2\frac{m}{s^2}$

$5000\frac{m}{s^2}$

$2500\pi \frac{m}{s^2}$

Correct answer:

$5000\pi^2\frac{m}{s^2}$

Explanation:

Centripetal acceleration $a_{cent}$ can be given as:

$a_{cent}=\frac{v^2}{r}$ , where is the linear velocity, is the radius of the circle.

r=.5m

To determine linear velocity, we need to determine how quickly the wheel spins.

Firstly, let's convert from rpm to Hertz.

$3000\frac{1}{min}*\frac{1min}{60s}= 50Hz$

The wheel makes 50 full revolutions in one second. To determine linear velocity,

v=f*C , where is frequency, which we just determined and is circumference of the tire.

By the definition of the circumference of a circle,

$C=2\pi r=2\pi*.5m=\pi m$

f=50Hz

$v=50Hz*\pi m= 50\pi \frac{m}{s}$

Therefore,

$a_{cent}=\frac{(50\pi\frac{m}{s})^2}{.5m}= 5000\pi^2 \frac{m}{s^2}$

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Example Question #51 : Circular And Rotational Motion

A ferris wheel ride of diameter 10m has a cart moving at $5\frac{m}{s}$ . What is the centripetal acceleration of the cart?

Possible Answers:

$0.2 \frac{m}{s^2}$

$1\frac{m}{s^2}$

$5\frac{m}{s^2}$

$2.5\frac{m}{s^2}$

Correct answer:

$5\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is given by:

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

$\alpha$ is the centripetal acceleration, is the velocity and is the radius of the circular path. We substitute in our known values and then solve for alpha which will give us the centripetal acceleration of the ferris wheel.

$\alpha= \frac{(5\frac{m}{s})^2}{5m}= \frac{25\frac{m^2}{s^2}}{5m}= 5\frac{m}{s^2}$

Be aware that the radius is half of the diameter, so will be 5 meters.

This will give us a centripetal acceleration of $5\frac{m}{s^2}$

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Example Question #51 : Circular And Rotational Motion

Vertical circle

A ball of mass $m=0.22\;kg$ is being swung in a vertical circle on a string, as shown in the given figure. At the instant it is directly above the center of the circle it is making, its speed is $v=2.7\frac{m}{s}$ . The string has negligible mass, and its length is $\l=0.61\,m$ . What is the tension, , in the string at the instant the ball is at the top of its circle?

Possible Answers:

$T=0\,N$

$T=2.16\,N$

$T=0.47\,N$

$T=4.79\,N$

$T=2.63\,N$

Correct answer:

$T=0.47\,N$

Explanation:

The tension and gravity both point down toward the center of the circle. The net force on the ball is the vector sum of these two:

$\Sigma F=T+mg$

For an object in circular motion, the net force must be equal to:

$\Sigma F=\frac{mv^2}{r}$

In this case, the force of gravity is helping the tension provide this force, so the tension is not as great as it would be if it were supplying the entire centripetal force.

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

Substitute and solve.

$T=\frac{mv^2}{r}-mg=\frac{0.22\times 2.1^2}{.61}-0.22\times9.8=0.47\,N$

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Example Question #56 : Circular And Rotational Motion

When a road is dry, a particular car can safely navigate a turn with a 50 m radius of curvature at $20 \frac{m}{s}$ without slipping. What is the coefficient of friction if this is the fastest speed the car can take this turn? Is this the static or kinetic coefficient?

Possible Answers:

0.5 , kinetic

0.04 , kinetic

0.82 , kinetic

0.04 , static

0.82 , static

Correct answer:

0.82 , static

Explanation:

For the car going around a flat turn, the force of friction is the only force keeping the car in a circular path. This lets us equate the centripetal force to the force of friction.

F_c=F_f

$\frac{mv^2}{r}=\mu N$

N=mg

Substitute and solve for the coefficient of friction.

$\mu=\frac{v^2}{gr}=\frac{(20\frac{m}{s})^2}{9.81\frac{m}{s^2}(50m)}=0.82$

Since the tires are rotating but not slipping on the surface, this is the coefficient of static friction, $\mu_s$ .

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Example Question #57 : Circular And Rotational Motion

On a wet road, the coefficient of static friction between a car's tires and the flat road is $\mu_s=0.4$ . What is the maximum speed a car can safely navigate a turn with a 30m radius of curvature?

Possible Answers:

$3.46 \frac{m}{s}$

$26.0 \frac{m}{s}$

$117.7 \frac{m}{s}$

$12.0 \frac{m}{s}$

$10.85 \frac{m}{s}$

Correct answer:

$10.85 \frac{m}{s}$

Explanation:

When a car is going around a flat turn, the only force keeping the car going along a circular path is the force of friction, which is equal to the centripetal force. We do not need the mass of the car to calculate the maximum speed, as we will show below.

F_c=F_f

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

N=mg

Rearrange these equations and solve for the velocity.

$v = \sqrt{\mu_s g r}$

$v= \sqrt{0.4(9.81\frac{m}{s^2})30m}=10.85\frac{m}{s}$

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Example Question #58 : Circular And Rotational Motion

If a car can safely go around an icy banked $\left(30^{\circ} \right )$ turn at a speed of $24 \frac{m}{s}$ , what is the radius of curvature $\left(r \right )$ of the turn?

Possible Answers:

117.4m

67.8m

Cannot be determined without knowing the mass of the car

4.24m

102m

Correct answer:

102m

Explanation:

For a car going around a banked turn with no friction, a component of the normal force is what provides the centripetal force. The radius of the curve can be calculated without knowing the mass of the car.

$mg\tan{\theta}=m\frac{v^2}{r}$

$r=\frac{v^2}{g\tan{\theta}}$

$r=\frac{\left(24\frac{m}{s} \right )^2}{9.81\frac{m}{s^2}\tan{30^{\circ}}}=102m$

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Example Question #59 : Circular And Rotational Motion

You tie a 1.2 kg rock to a rope and swing it above your head with a frequency of $\frac{1.1rotations}{ second}$ . The distance from your hand to the rope is .45 meters . Determine the centripital force experience by the rock.

Possible Answers:

9.1 N

31.3 N

17.6 N

None of these

12.2 N

Correct answer:

17.6 N

Explanation:

Centripital acceleration is

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

Force is defined as

F=ma

Using subsitution

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

The rock has a frequency of $\frac{1.1rotations}{ second}$ .

Each circle it makes has a circumference of $2\cdot\pi\cdot .45meters$ , which is equivalent to 2.82 meters .

That means it travel 2.82 meters in 1.1 seconds , which is $2.57 \frac{meters}{second}$

Plugging in our values:

$F=1.2\frac{2.57^2}{.45}$

F=17.6 N

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Example Question #60 : Circular And Rotational Motion

A ball attached to a 1 meter-long string is being spun around and is traveling at $4\frac{m}{s}$ .

What is the centripetal acceleration of the ball?

Possible Answers:

$4\frac{m}{s^2}$

$16\frac{m}{s^2}$

$8\frac{m}{s^2}$

$\frac{1}{4}\frac{m}{s^2}$

$\frac{1}{16}\frac{m}{s^2}$

Correct answer:

$16\frac{m}{s^2}$

Explanation:

We use the formula for centripetal acceleration:

$a_{c}=\frac{v^{2}}{r}$

Where a_c is the centripetal acceleration, is the velocity $\left(4\frac{m}{s}\right)$ and is the radius (1m).

Plug in these known values and solve.

$a_c=\frac{(4^2\frac{m}{s})}{1m}$

$a_c=16\frac{m}{s^2}$

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AP Physics 1 : Circular and Rotational Motion

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #51 : Circular And Rotational Motion

Example Question #51 : Circular And Rotational Motion

Example Question #53 : Circular And Rotational Motion

Example Question #51 : Circular And Rotational Motion

Example Question #51 : Circular And Rotational Motion

Example Question #56 : Circular And Rotational Motion

Example Question #57 : Circular And Rotational Motion

Example Question #58 : Circular And Rotational Motion

Example Question #59 : Circular And Rotational Motion

Example Question #60 : Circular And Rotational Motion

All AP Physics 1 Resources

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