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High School Physics : Motion and Mechanics

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Example Questions

Example Question #2 : 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 force?

Possible Answers:

98N $\displaystyle 98N$

64N $\displaystyle 64N$

256N $\displaystyle 256N$

128N $\displaystyle 128N$

$\displaystyle 8N$

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 $\displaystyle F_c=ma_c$.

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.

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

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

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

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 #4 : Using Circular Motion Equations

Possible Answers:

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

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

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

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

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

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.

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.

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

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

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

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

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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 #3 : Using Circular Motion Equations

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 $\displaystyle 7.5N*m$

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

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

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

$\displaystyle 75000N*m$

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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Example Question #11 : Using Circular Motion Equations

The velocity of a certain ferris wheel is $6\frac{m}{s}$ $\displaystyle 6\frac{m}{s}$. If the wheel has a diameter of 12m $\displaystyle 12m$, what is the centripetal acceleration generated by the wheel?

Possible Answers:

$30\frac{m}{s^2}$ $\displaystyle 30\frac{m}{s^2}$

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

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

$36\frac{m}{s^2}$ $\displaystyle 36\frac{m}{s^2}$

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

Correct answer:

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

Explanation:

The formula for centripetal acceleration is $a_c=\frac{v^2}{r}$ $\displaystyle a_c=\frac{v^2}{r}$.

The problem gives us a diamater. Since radius is defined as half of an object's diameter, the radius of the ferris wheel is $\displaystyle 6m$.

Now we can plug in and solve:

$a_c=\frac{v^2}{r}$ $\displaystyle a_c=\frac{v^2}{r}$

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

$a_c=\frac{36\frac{m^2}{s^2}}{6m}$ $\displaystyle a_c=\frac{36\frac{m^2}{s^2}}{6m}$

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

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Example Question #11 : Using Circular Motion Equations

An object moves in a circle with a constant velocity of $3\frac{m}{s}$ $\displaystyle 3\frac{m}{s}$. If the radius of the circle is $\displaystyle 2m$, what is the centripetal acceleration on the object?

Possible Answers:

$4.5\frac{m}{s^2}$ $\displaystyle 4.5\frac{m}{s^2}$

$18\frac{m}{s^2}$ $\displaystyle 18\frac{m}{s^2}$

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

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

$1.5\frac{m}{s^2}$ $\displaystyle 1.5\frac{m}{s^2}$

Correct answer:

$4.5\frac{m}{s^2}$ $\displaystyle 4.5\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is:

$a_c=\frac{v^2}{r}$ $\displaystyle a_c=\frac{v^2}{r}$

We are given the velocity and the radius, allowing us to solve for the acceleration.

$a_c=\frac{v^2}{r}$ $\displaystyle a_c=\frac{v^2}{r}$

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

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

$a_c=4.5\frac{m}{s^2}$ $\displaystyle a_c=4.5\frac{m}{s^2}$

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Example Question #11 : Using Circular Motion Equations

If the centripetal force on a 3.3kg $\displaystyle 3.3kg$ object is 14.5N $\displaystyle 14.5N$, and the object is 1.9m $\displaystyle 1.9m$ from the center of the circle, what is its centripetal acceleration?

Possible Answers:

$4.43\frac{m}{s^2}$ $\displaystyle 4.43\frac{m}{s^2}$

$3.3\frac{m}{s^2}$ $\displaystyle 3.3\frac{m}{s^2}$

$11.2\frac{m}{s^2}$ $\displaystyle 11.2\frac{m}{s^2}$

$0.443\frac{m}{s^2}$ $\displaystyle 0.443\frac{m}{s^2}$

$5.6\frac{m}{s^2}$ $\displaystyle 5.6\frac{m}{s^2}$

Correct answer:

$4.43\frac{m}{s^2}$ $\displaystyle 4.43\frac{m}{s^2}$

Explanation:

The formula for centripetal force is $\displaystyle F_c=ma_c$.

We know the value of the force, as well as the mass of the object. Using these values, we can solve for the acceleration; the radius is extraneous information.

$\displaystyle F_c=ma_c$

$\displaystyle 14.5N=3.3kg*a_c$

$\frac{14.5N}{3.3kg}=a_c$ $\displaystyle \frac{14.5N}{3.3kg}=a_c$

$4.34\frac{m}{s^2}=a_c$ $\displaystyle 4.34\frac{m}{s^2}=a_c$

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Example Question #12 : Using Circular Motion Equations

Possible Answers:

$4.34\frac{m}{s}$ $\displaystyle 4.34\frac{m}{s}$

$8.25\frac{m}{s}$ $\displaystyle 8.25\frac{m}{s}$

$2.083\frac{m}{s}$ $\displaystyle 2.083\frac{m}{s}$

$1.51\frac{m}{s}$ $\displaystyle 1.51\frac{m}{s}$

$2.87\frac{m}{s}$ $\displaystyle 2.87\frac{m}{s}$

Correct answer:

$2.87\frac{m}{s}$ $\displaystyle 2.87\frac{m}{s}$

Explanation:

The relationship between centripetal acceleration and tangential velocity is:

$a_c=\frac{v^2}{r}$ $\displaystyle a_c=\frac{v^2}{r}$

We can find the centripetal acceleration using the centripetal force and the mass.

The formula for centripetal force is $\displaystyle F_c=ma_c$.

Using the force and mass from the question, we can solve for the acceleration.

$\displaystyle F_c=ma_c$

$\displaystyle 14.5N=3.3kg*a_c$

$\frac{14.5N}{3.3kg}=a_c$ $\displaystyle \frac{14.5N}{3.3kg}=a_c$

$4.34\frac{m}{s^2}=a_c$ $\displaystyle 4.34\frac{m}{s^2}=a_c$

Now that we know the acceleration, we can return to the first equation. Using the acceleration and the radius from the question, we can solve for the velocity.

$a_c=\frac{v^2}{r}$ $\displaystyle a_c=\frac{v^2}{r}$

$4.34\frac{m}{s^2}=\frac{v^2}{1.9m}$ $\displaystyle 4.34\frac{m}{s^2}=\frac{v^2}{1.9m}$

$4.34\frac{m}{s^2}*1.9m=v^2$ $\displaystyle 4.34\frac{m}{s^2}*1.9m=v^2$

$8.246\frac{m^2}{s^2}=v^2$ $\displaystyle 8.246\frac{m^2}{s^2}=v^2$

$\sqrt{8.246\frac{m^2}{s^2}}=\sqrt{v^2}$ $\displaystyle \sqrt{8.246\frac{m^2}{s^2}}=\sqrt{v^2}$

$2.87\frac{m}{s}=v$ $\displaystyle 2.87\frac{m}{s}=v$

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High School Physics : Motion and Mechanics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #2 : Using Circular Motion Equations

Example Question #4 : Using Circular Motion Equations

Example Question #7 : Using Circular Motion Equations

Example Question #8 : Using Circular Motion Equations

Example Question #3 : Using Circular Motion Equations

Example Question #1 : Using Circular Motion Equations

Example Question #11 : Using Circular Motion Equations

Example Question #11 : Using Circular Motion Equations

Example Question #11 : Using Circular Motion Equations

Example Question #12 : Using Circular Motion Equations

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