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High School Physics : Using Circular Motion Equations

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Example Questions

Example Question #12 : Circular Motion

A 2kg ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $3\frac{m}{s}$ , what is the centripetal acceleration?

Possible Answers:

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

$3\frac{m}{s^2}$

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

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

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

Correct answer:

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

Explanation:

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(3\frac{m}{s})^2}{(1m)}$

$a_c=\frac{9\frac{m^2}{s^2}}{1m}$

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

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Example Question #222 : High School Physics

A 2kg ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $3\frac{m}{s}$ , what is the centripetal force?

Possible Answers:

$9\pi N$

18N

Correct answer:

18N

Explanation:

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is F_c=ma_c .

To find the centripetal force, we need to find the centripetal acceleration. We do this with the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(3\frac{m}{s})^2}{(1m)}$

$a_c=\frac{9\frac{m^2}{s^2}}{1m}$

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

Plug that into the first equation to solve for the force.

$F_c=(2kg)(9\frac{m}{s^2})$

F_c=18N

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Example Question #1 : Using Circular Motion Equations

A 2kg ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of $3\frac{m}{s}$ , what is the period of the ball?

Possible Answers:

$\pi s$

$2\pi s$

$\frac{3\pi}{2}s$

$\frac{2\pi}{3}s$

$\frac{\pi}{2}s$

Correct answer:

$\frac{2\pi}{3}s$

Explanation:

The period, , of an object moving in circular motion is the amount of time it takes for the object to make one complete loop of the circle.

If we start with the linear understanding of velocity, $v=\frac{\Delta x}{\Delta t}$ , we can apply the same concept here. Our velocity should be the change in distance over the change in time. In this case, we don't have a definite time, , but we do have a period in terms of one complete loop.

We can set up an equation for the period using the circumference of the circle as our distance: $v=\frac{2\pi r}{T}$ .

Plug in the given values to solve for the period.

$v=\frac{2\pi r}{T}$

$3\frac{m}{s}=\frac{2\pi (1m)}{T}$

$T=\frac{2\pi (1m)}{3\frac{m}{s}}$

$T=\frac{2\pi}{3}s$

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Example Question #15 : Circular Motion

A 2kg ball rolls around the edge of a circle with a radius of 0.5m . If it is rolling at a speed of $8\frac{m}{s}$ , what is the centripetal acceleration?

Possible Answers:

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

$128\frac{m}{s^2}$

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

$64\frac{m}{s^2}$

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

Correct answer:

$128\frac{m}{s^2}$

Explanation:

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

$a_c=\frac{(8\frac{m}{s})^2}{(0.5m)}$

$a_c=\frac{64\frac{m^2}{s^2}}{0.5m}$

$a_c=128\frac{m}{s^2}$

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Example Question #2 : Using Circular Motion Equations

A 2kg ball rolls around the edge of a circle with a radius of 0.5m . If it is rolling at a speed of $8\frac{m}{s}$ , what is the centripetal force?

Possible Answers:

98N

256N

128N

64N

Correct answer:

256N

Explanation:

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is F_c=ma_c .

Plug in the given values and solve for the acceleration.

$a_c=\frac{(8\frac{m}{s})^2}{(0.5m)}$

$a_c=\frac{64\frac{m^2}{s^2}}{0.5m}$

$a_c=128\frac{m}{s^2}$

Plug the acceleration and given mass into the first equation to solve for force.

$F_c=(2kg)(128\frac{m}{s^2})$

F_c=256N

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Example Question #224 : High School Physics

A 2kg ball rolls around the edge of a circle with a radius of 0.5m . If it is rolling at a speed of $8\frac{m}{s}$ , what is the period of the ball?

Possible Answers:

$16\pi s$

$\frac{\pi}{4}s$

$\frac{\pi}{8}s$

$\frac{2\pi}{3}s$

$8\pi s$

Correct answer:

$\frac{\pi}{8}s$

Explanation:

The period, , of an object moving in circular motion is the amount of time it takes for the object to make one complete loop of the circle.

We can set up an equation for the period using the circumference of the circle as our distance: $v=\frac{2\pi r}{T}$ .

Plug in the given values to solve for the period.

$v=\frac{2\pi r}{T}$

$8\frac{m}{s}=\frac{2\pi (0.5m)}{T}$

$T=\frac{\pi m}{8\frac{m}{s}}$

$T=\frac{\pi}{8}s$

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Example Question #1 : Using Circular Motion Equations

A 2kg ball rolls around the edge of a circle with a radius of . If there is a 12N centripital force acting upon it, what is its speed?

Possible Answers:

$3.99\frac{m}{s}$

$2.45\frac{m}{s}$

$6\frac{m}{s}$

$3\frac{m}{s}$

$1.82\frac{m}{s}$

Correct answer:

$2.45\frac{m}{s}$

Explanation:

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is F_c=ma_c .

Since we know the mass and the force, we can find the accleration.

F_c=ma_c

12N=(2kg)a_c

$\frac{12N}{2kg}=a_c$

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

Centripetal acceleration is given by the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the velocity.

$6\frac{m}{s^2}=\frac{v^2}{(1m)}$

$1m* 6\frac{m}{s^2}=v^2$

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

$\sqrt{6\frac{m^2}{s^2}}=\sqrt{v^2}$

$2.45\frac{m}{s}=v$

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Example Question #3 : Using Circular Motion Equations

The radius of the sun is 696,000km. If its period is 587.28 hours, what is its tangential velocity at the equator?

Possible Answers:

$2370.25\pi\frac{km}{hr}$

$700000\pi\frac{km}{hr}$

$98101.23\frac{km}{hr}$

$14446.36\pi\frac{km}{hr}$

$589.22\frac{km}{hr}$

Correct answer:

$2370.25\pi\frac{km}{hr}$

Explanation:

The equation for velocity is $v=\frac{\Delta x}{\Delta t}$ . Using the circumference of the circle as the distance and the time as the period, we can rewrite the equation for velocity: $v=\frac{2\pi r}{T}$ .

Plug in the given values and solve for the velocity.

$v=\frac{2\pi (696,000km)}{587.28hr}$

$v=\frac{\pi*1392000km}{587.28hr}$

$v=2370.25\pi\frac{km}{hr}$

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Example Question #1 : Using Circular Motion Equations

A large burst of wind exerts 750N of force on a tree 10m tall. How much torque is exerted on the tree? (Assume that the entire force is pointed at the top of the tree, the pivot point is at the ground, and ignore center of mass fluctuations)

Possible Answers:

740N*m

7500N*m

75000N*m

7.5N*m

75N*m

Correct answer:

7500N*m

Explanation:

The formula for torque is:

$\tau=F*r$

Plug in our given values:

$\tau=750N*10m$

$\tau=7500N*m$

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Example Question #1 : Using Circular Motion Equations

of force is exerted on the end of a 0.23m wrench. How much torque is generated? (Ignore any center of mass fluctuations.)

Possible Answers:

0.92N*m

7.77N*m

8.23N*m

34.78N*m

1.84N*m

Correct answer:

1.84N*m

Explanation:

The formula for torque is:

$\tau=F*r$

Plug in our given values:

$\tau=8N*0.23m$

$\tau=1.84N*m$

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High School Physics : Using Circular Motion Equations

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #12 : Circular Motion

Example Question #222 : High School Physics

Example Question #1 : Using Circular Motion Equations

Example Question #15 : Circular Motion

Example Question #2 : Using Circular Motion Equations

Example Question #224 : High School Physics

Example Question #1 : Using Circular Motion Equations

Example Question #3 : Using Circular Motion Equations

Example Question #1 : Using Circular Motion Equations

Example Question #1 : Using Circular Motion Equations

All High School Physics Resources

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