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High School Physics : Using Circular Motion Equations

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Example Questions

Example Question #14 : Circular Motion

A 2kg $\displaystyle 2kg$ ball rolls around the edge of a circle with a radius of $\displaystyle 1m$. If it is rolling at a speed of $3\frac{m}{s}$ $\displaystyle 3\frac{m}{s}$, what is the centripetal acceleration?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

Plug in the given values and solve for the acceleration.

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

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

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

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Example Question #1 : Using Circular Motion Equations

Possible Answers:

$\displaystyle 2N$

$\displaystyle 1N$

$\displaystyle 9N$

18N $\displaystyle 18N$

$9\pi N$ $\displaystyle 9\pi N$

Correct answer:

18N $\displaystyle 18N$

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 $\displaystyle F_c=ma_c$.

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

Plug in the given values and solve for the acceleration.

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

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

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

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

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

F_c=18N $\displaystyle F_c=18N$

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Example Question #11 : Circular Motion

Possible Answers:

$\pi s$ $\displaystyle \pi s$

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

$2\pi s$ $\displaystyle 2\pi s$

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

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

Correct answer:

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

Explanation:

The period, $\displaystyle T$, 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, $v=\frac{\Delta x}{\Delta t}$ $\displaystyle v=\frac{\Delta x}{\Delta t}$, 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, $\displaystyle t$, 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: $v=\frac{2\pi r}{T}$ $\displaystyle v=\frac{2\pi r}{T}$.

Plug in the given values to solve for the period.

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

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

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

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

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Example Question #4 : Using Circular Motion Equations

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

Plug in the given values and solve for the acceleration.

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

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

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

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Example Question #222 : High School Physics

Possible Answers:

$\displaystyle 8N$

128N $\displaystyle 128N$

98N $\displaystyle 98N$

256N $\displaystyle 256N$

64N $\displaystyle 64N$

Correct answer:

256N $\displaystyle 256N$

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 $\displaystyle F_c=ma_c$.

Plug in the given values and solve for the acceleration.

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

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

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

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

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

$\displaystyle F_c=256N$

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Example Question #1 : Using Circular Motion Equations

Possible Answers:

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

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

$8\pi s$ $\displaystyle 8\pi s$

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

$16\pi s$ $\displaystyle 16\pi s$

Correct answer:

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

Explanation:

The period, $\displaystyle T$, 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: $v=\frac{2\pi r}{T}$ $\displaystyle v=\frac{2\pi r}{T}$.

Plug in the given values to solve for the period.

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

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

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

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

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Example Question #7 : Using Circular Motion Equations

A 2kg $\displaystyle 2kg$ ball rolls around the edge of a circle with a radius of $\displaystyle 1m$. If there is a 12N $\displaystyle 12N$ centripital force acting upon it, what is its speed?

Possible Answers:

$3\frac{m}{s}$ $\displaystyle 3\frac{m}{s}$

$2.45\frac{m}{s}$ $\displaystyle 2.45\frac{m}{s}$

$1.82\frac{m}{s}$ $\displaystyle 1.82\frac{m}{s}$

$3.99\frac{m}{s}$ $\displaystyle 3.99\frac{m}{s}$

$6\frac{m}{s}$ $\displaystyle 6\frac{m}{s}$

Correct answer:

$2.45\frac{m}{s}$ $\displaystyle 2.45\frac{m}{s}$

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 $\displaystyle F_c=ma_c$.

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

$\displaystyle F_c=ma_c$

$\displaystyle 12N=(2kg)a_c$

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

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

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

Plug in the given values and solve for the velocity.

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

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

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

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

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

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Example Question #8 : 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:

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

$98101.23\frac{km}{hr}$ $\displaystyle 98101.23\frac{km}{hr}$

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

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

$589.22\frac{km}{hr}$ $\displaystyle 589.22\frac{km}{hr}$

Correct answer:

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

Explanation:

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

Plug in the given values and solve for the velocity.

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

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

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

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Example Question #231 : High School Physics

A large burst of wind exerts 750N $\displaystyle 750N$ of force on a tree 10m $\displaystyle 10m$ 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:

7.5N*m $\displaystyle 7.5N*m$

7500N*m $\displaystyle 7500N*m$

75N*m $\displaystyle 75N*m$

740N*m $\displaystyle 740N*m$

$\displaystyle 75000N*m$

Correct answer:

7500N*m $\displaystyle 7500N*m$

Explanation:

The formula for torque is:

$\tau=F*r$ $\displaystyle \tau=F*r$

Plug in our given values:

$\tau=750N*10m$ $\displaystyle \tau=750N*10m$

$\tau=7500N*m$ $\displaystyle \tau=7500N*m$

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Example Question #1 : Using Circular Motion Equations

$\displaystyle 8N$ of force is exerted on the end of a 0.23m $\displaystyle 0.23m$ wrench. How much torque is generated? (Ignore any center of mass fluctuations.)

Possible Answers:

$\displaystyle 34.78N*m$

0.92N*m $\displaystyle 0.92N*m$

7.77N*m $\displaystyle 7.77N*m$

8.23N*m $\displaystyle 8.23N*m$

1.84N*m $\displaystyle 1.84N*m$

Correct answer:

1.84N*m $\displaystyle 1.84N*m$

Explanation:

The formula for torque is:

$\tau=F*r$ $\displaystyle \tau=F*r$

Plug in our given values:

$\tau=8N*0.23m$ $\displaystyle \tau=8N*0.23m$

$\tau=1.84N*m$ $\displaystyle \tau=1.84N*m$

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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 #14 : Circular Motion

Example Question #1 : Using Circular Motion Equations

Example Question #11 : Circular Motion

Example Question #4 : Using Circular Motion Equations

Example Question #222 : High School Physics

Example Question #1 : Using Circular Motion Equations

Example Question #7 : Using Circular Motion Equations

Example Question #8 : Using Circular Motion Equations

Example Question #231 : High School Physics

Example Question #1 : Using Circular Motion Equations

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