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High School Physics : Circular Motion

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Example Questions

Example Question #51 : Circular Motion

A 2.2kg $\displaystyle 2.2kg$ object is moving in a perfect circle with a radius of 3.1m $\displaystyle 3.1m$ and has a linear momentum of $35\frac{kg*m}{s}$ $\displaystyle 35\frac{kg*m}{s}$ .

What is the angular momentum of the object?

Possible Answers:

$57.9\,J*s$ $\displaystyle 57.9\,J*s$

$179.63\,J*s$ $\displaystyle 179.63\,J*s$

$11.29\,J*s$ $\displaystyle 11.29\,J*s$

$5.13\,J*s$ $\displaystyle 5.13\,J*s$

$108.5\,J*s$ $\displaystyle 108.5\,J*s$

Correct answer:

$108.5\,J*s$ $\displaystyle 108.5\,J*s$

Explanation:

There is a direct relationship between angular momentum and linear momentum. Angular momentum is equal to the linear momentum times the radius:

L=p*r $\displaystyle L=p*r$

We are given the value of both the linear momentum and the radius, allowing us to solve for the angular momentum.

$L=(35kg*\frac{m}{s})(3.1m)$ $\displaystyle L=(35kg*\frac{m}{s})(3.1m)$

$L=108.5\,J*s$ $\displaystyle L=108.5\,J*s$

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Example Question #52 : Circular Motion

A 2.2kg $\displaystyle 2.2kg$ object is moving in a perfect circle with a radius of 3.1m $\displaystyle 3.1m$ and has a linear momentum of $35\frac{kg*m}{s}$ $\displaystyle 35\frac{kg*m}{s}$ .

What is the moment of inertia on this object?

Possible Answers:

$\displaystyle 9.61kg*m^2$

$\displaystyle 5.13kg*m^2$

$\displaystyle 15kg*m^2$

$\displaystyle 6.82kg*m^2$

$\displaystyle 21.14kg*m^2$

Correct answer:

$\displaystyle 21.14kg*m^2$

Explanation:

Since the object is moving in a perfect circle and not rotating about some fixed point within itself (like spinning a ball or a frisbee), the equation for moment of inertia is:

I=mr^2 $\displaystyle I=mr^2$

We are given the mass and radius, allowing us to calculate the moment of inertia from this equation.

$\displaystyle I=(2.2kg)(3.1m)^2$

$\displaystyle I=(2.2kg)(9.61m^2)$

$\displaystyle I=21.14kg*m^2$

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Example Question #53 : Circular Motion

A 2.2kg $\displaystyle 2.2kg$ object is moving in a perfect circle with a radius of 3.1m $\displaystyle 3.1m$ and has a linear momentum of $35\frac{kg*m}{s}$ $\displaystyle 35\frac{kg*m}{s}$ .

What is the period of the object's orbit?

Possible Answers:

1.5s $\displaystyle 1.5s$

1.22s $\displaystyle 1.22s$

0.61s $\displaystyle 0.61s$

0.82s $\displaystyle 0.82s$

0.39s $\displaystyle 0.39s$

Correct answer:

1.22s $\displaystyle 1.22s$

Explanation:

The relationship between period and angular velocity is:

$\omega=\frac{2\pi}{T}$ $\displaystyle \omega=\frac{2\pi}{T}$

The relationship between linear and angular velocity is:

$\omega=\frac{v}{r}$ $\displaystyle \omega=\frac{v}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

p=m*v $\displaystyle p=m*v$

$35\frac{kg*m}{s}=(2.2kg)v$ $\displaystyle 35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$ $\displaystyle \frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$ $\displaystyle 15.91\frac{m}{s}=v$

Plug that into the equation for angular velocity, along with the radius.

$\omega=\frac{15.91\frac{m}{s}}{3.1m}$ $\displaystyle \omega=\frac{15.91\frac{m}{s}}{3.1m}$

$\omega=5.13\frac{rad}{s}$ $\displaystyle \omega=5.13\frac{rad}{s}$

Plug this term back into the initial equation to solve for the period.

$\omega=\frac{2\pi}{T}$ $\displaystyle \omega=\frac{2\pi}{T}$

$5.13\frac{rad}{s}=\frac{2\pi}{T}$ $\displaystyle 5.13\frac{rad}{s}=\frac{2\pi}{T}$

$T=\frac{2\pi}{5.13\frac{rad}{s}}$ $\displaystyle T=\frac{2\pi}{5.13\frac{rad}{s}}$

T=1.22s $\displaystyle T=1.22s$

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Example Question #191 : Motion And Mechanics

A 2.2kg $\displaystyle 2.2kg$ object is moving in a perfect circle with a radius of 3.1m $\displaystyle 3.1m$ and has a linear momentum of $35\frac{kg*m}{s}$ $\displaystyle 35\frac{kg*m}{s}$ .

What is the angular displacement of the object after $\displaystyle 5s$ ?

Possible Answers:

$\displaystyle 79.55rad$

$\displaystyle 161.16rad$

$\displaystyle 26.65rad$

9234rad $\displaystyle 9234rad$

4.08rad $\displaystyle 4.08rad$

Correct answer:

$\displaystyle 26.65rad$

Explanation:

Angular displacement is equal to the angular velocity times time:

$\theta=\omega*\Delta t$ $\displaystyle \theta=\omega*\Delta t$

We know the time, but we need to solve for the angular velocity. The relationship between linear and angular velocity is:

$\omega=\frac{v}{r}$ $\displaystyle \omega=\frac{v}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

p=m*v $\displaystyle p=m*v$

$35\frac{kg*m}{s}=(2.2kg)v$ $\displaystyle 35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$ $\displaystyle \frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$ $\displaystyle 15.91\frac{m}{s}=v$

Plug that into the equation for angular velocity, along with the radius.

$\omega=\frac{15.91\frac{m}{s}}{3.1m}$ $\displaystyle \omega=\frac{15.91\frac{m}{s}}{3.1m}$

$\omega=5.13\frac{rad}{s}$ $\displaystyle \omega=5.13\frac{rad}{s}$

Now that we have the angular velocity, we can solve for the angular displacement at the given time.

$\theta=\omega *\Delta t$ $\displaystyle \theta=\omega *\Delta t$

$\theta=5.13\frac{rad}{s}*5s$ $\displaystyle \theta=5.13\frac{rad}{s}*5s$

$\theta=26.65rad$ $\displaystyle \theta=26.65rad$

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Example Question #32 : Using Circular Motion Equations

A 2.2kg $\displaystyle 2.2kg$ object is moving in a perfect circle with a radius of 3.1m $\displaystyle 3.1m$ and has a linear momentum of $35\frac{kg*m}{s}$ $\displaystyle 35\frac{kg*m}{s}$ .

How many revolutions does the object go through in 14s $\displaystyle 14s$ ?

Possible Answers:

$17.08\, rev$ $\displaystyle 17.08\, rev$

$11.48\, rev$ $\displaystyle 11.48\, rev$

$0.09\, rev$ $\displaystyle 0.09\, rev$

$72.1 \, rev$ $\displaystyle 72.1 \, rev$

$1.22\, rev$ $\displaystyle 1.22\, rev$

Correct answer:

$11.48\, rev$ $\displaystyle 11.48\, rev$

Explanation:

We can set up a proportion here:

$\frac{1rev}{T}=\frac{x\, rev}{14s}$ $\displaystyle \frac{1rev}{T}=\frac{x\, rev}{14s}$ .

In other words, the object can do one revolution per period ( $\displaystyle T$ ), so it can do $\displaystyle x$ revolutions in 14s $\displaystyle 14s$ . First we need to find the period.

The relationship between period and angular velocity is:

$\omega=\frac{2\pi}{T}$ $\displaystyle \omega=\frac{2\pi}{T}$

The relationship between linear and angular velocity is:

$\omega=\frac{v}{r}$ $\displaystyle \omega=\frac{v}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

p=m*v $\displaystyle p=m*v$

$35\frac{kg*m}{s}=(2.2kg)v$ $\displaystyle 35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$ $\displaystyle \frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$ $\displaystyle 15.91\frac{m}{s}=v$

Plug that into the equation for angular velocity, along with the radius.

$\omega=\frac{15.91\frac{m}{s}}{3.1m}$ $\displaystyle \omega=\frac{15.91\frac{m}{s}}{3.1m}$

$\omega=5.13\frac{rad}{s}$ $\displaystyle \omega=5.13\frac{rad}{s}$

Plug this term back into the first equation to solve for the period.

$\omega=\frac{2\pi}{T}$ $\displaystyle \omega=\frac{2\pi}{T}$

$5.13\frac{rad}{s}=\frac{2\pi}{T}$ $\displaystyle 5.13\frac{rad}{s}=\frac{2\pi}{T}$

$T=\frac{2\pi}{5.13\frac{rad}{s}}$ $\displaystyle T=\frac{2\pi}{5.13\frac{rad}{s}}$

T=1.22s $\displaystyle T=1.22s$

Now that we know the period, we can return to the proportion to solve for the rotations completed in the given time.

$\frac{1rev}{1.22s}=\frac{x\, rev}{14s}$ $\displaystyle \frac{1rev}{1.22s}=\frac{x\, rev}{14s}$

Cross multiply.

$14\, rev*s=(1.22s)(x)$ $\displaystyle 14\, rev*s=(1.22s)(x)$

$\frac{14\,rev*s}{1.22s}=x$ $\displaystyle \frac{14\,rev*s}{1.22s}=x$

$11.48\, rev = x$ $\displaystyle 11.48\, rev = x$

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Example Question #56 : Circular Motion

A 2.2kg $\displaystyle 2.2kg$ object is moving in a perfect circle with a radius of 3.1m $\displaystyle 3.1m$ and has a linear momentum of $35\frac{kg*m}{s}$ $\displaystyle 35\frac{kg*m}{s}$ .

What is the centripetal force on the object?

Possible Answers:

179.63N $\displaystyle 179.63N$

395.19N $\displaystyle 395.19N$

15.91N $\displaystyle 15.91N$

81.65N $\displaystyle 81.65N$

129.36N $\displaystyle 129.36N$

Correct answer:

179.63N $\displaystyle 179.63N$

Explanation:

Newton's second law states that F=ma $\displaystyle F=ma$ . We know the mass of the object, but we need to find the centripetal acceleration to calculate the centripetal force.

Centripetal acceleration is equal to the tangential velocity squared over the radius:

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

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

p=m*v $\displaystyle p=m*v$

$35\frac{kg*m}{s}=(2.2kg)v$ $\displaystyle 35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$ $\displaystyle \frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$ $\displaystyle 15.91\frac{m}{s}=v$

Use the linear velocity and the radius in the previous equation to solve for the centripetal acceleration.

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

$a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$ $\displaystyle a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$

$a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$ $\displaystyle a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$

$a_c=81.65\frac{m}{s^2}$ $\displaystyle a_c=81.65\frac{m}{s^2}$

Use the centripetal acceleration in Newton's second law, along with the mass, to calculate the centripetal force.

$\displaystyle F_c=ma_c$

$F_c=2.2kg*81.65\frac{m}{s^2}$ $\displaystyle F_c=2.2kg*81.65\frac{m}{s^2}$

$\displaystyle F_c=179.63N$

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Example Question #57 : Circular Motion

A baseball has an angular velocity of $88\frac{rad}{s}$ $\displaystyle 88\frac{rad}{s}$ . If the baseball reaches the plate 0.45s $\displaystyle 0.45s$ after the pitcher releases it, what is the angular displacement of the ball?

Possible Answers:

$44\: rad$ $\displaystyle 44\: rad$

$39.6\: rad$ $\displaystyle 39.6\: rad$

$248.8\: rad$ $\displaystyle 248.8\: rad$

$2.83\: rad$ $\displaystyle 2.83\: rad$

Correct answer:

$39.6\: rad$ $\displaystyle 39.6\: rad$

Explanation:

The formula for angular displacement with no angular acceleration is $\theta=\omega*t$ $\displaystyle \theta=\omega*t$ .

Plug in our given values.

$\theta=\omega*t$ $\displaystyle \theta=\omega*t$

$\theta=88\frac{rads}{s}*0.45s$ $\displaystyle \theta=88\frac{rads}{s}*0.45s$

$\theta =39.6\: rad$ $\displaystyle \theta =39.6\: rad$

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Example Question #58 : Circular Motion

A baseball batter swings and misses at a baseball. If the length from his shoulder to the end of the bat is 1.6m $\displaystyle 1.6m$ , the bat has a mass of 1.1kg $\displaystyle 1.1kg$ , and the bat moves with a linear velocity of $18.3\frac{m}{s}$ $\displaystyle 18.3\frac{m}{s}$ , what is the angular momentum generated?

Possible Answers:

$\displaystyle 10.39J*s$

$\displaystyle 32.21J*s$

21.0J*s $\displaystyle 21.0J*s$

0.03J*s $\displaystyle 0.03J*s$

$\displaystyle 16.54J*s$

Correct answer:

$\displaystyle 32.21J*s$

Explanation:

Angular momentum is the product of linear momentum times the lever arm:

L=p*r $\displaystyle L=p*r$

Expand this equation by using the formula for linear momentum.

p=mv $\displaystyle p=mv$

L=mv*r $\displaystyle L=mv*r$

We are given the length of the lever arm (radius), the mass of the bat, and the linear velocity of the swing. Using these values, we can solve for the angular momentum.

L=mv*r $\displaystyle L=mv*r$

$L=(1.1kg)(18.3\frac{m}{s})(1.6m)$ $\displaystyle L=(1.1kg)(18.3\frac{m}{s})(1.6m)$

$\displaystyle L=32.21J*s$

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Example Question #59 : Circular Motion

A child swings a pail of water in a circle, using her shoulder as the pivot point. If the child's arm is 0.8m $\displaystyle 0.8m$ long and the pail of water has a linear momentum of $1.67\frac{kg\cdot m}{s}$ $\displaystyle 1.67\frac{kg\cdot m}{s}$ , what is the angular momentum of the pail of water?

Possible Answers:

$0.87\ \frac{kg\cdot m^2}{s}$ $\displaystyle 0.87\ \frac{kg\cdot m^2}{s}$

$2.23\ \frac{kg\cdot m^2}{s}$ $\displaystyle 2.23\ \frac{kg\cdot m^2}{s}$

$1.07\ \frac{kg\cdot m^2}{s}$ $\displaystyle 1.07\ \frac{kg\cdot m^2}{s}$

$1.3\ \frac{kg\cdot m^2}{s}$ $\displaystyle 1.3\ \frac{kg\cdot m^2}{s}$

$2.1\ \frac{kg\cdot m^2}{s}$ $\displaystyle 2.1\ \frac{kg\cdot m^2}{s}$

Correct answer:

$1.3\ \frac{kg\cdot m^2}{s}$ $\displaystyle 1.3\ \frac{kg\cdot m^2}{s}$

Explanation:

Angular momentum is equal to the linear momentum times the radial arm.

L=r*p $\displaystyle L=r*p$

Since the pivot point of the child is the shoulder, the radial arm is the child's arm length. We are given this length, as well as the linear momentum, allowing us to solve for the angular momentum.

$L=(0.8m)(1.67\frac{kg\cdot m}{s})$ $\displaystyle L=(0.8m)(1.67\frac{kg\cdot m}{s})$

$L=1.3\frac{kg\cdot m^2}{s}$ $\displaystyle L=1.3\frac{kg\cdot m^2}{s}$

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Example Question #60 : Circular Motion

A 0.02kg $\displaystyle 0.02kg$ disc has a radius of 0.11m $\displaystyle 0.11m$ and is thrown horizontally. If it has an angular velocity of $3.2\frac{rad}{s}$ $\displaystyle 3.2\frac{rad}{s}$ , what is the angular momentum of the disc?

$I_{disc}=\frac{1}{2}MR^2$ $\displaystyle I_{disc}=\frac{1}{2}MR^2$

Possible Answers:

$7.04*10^{-5}\ \frac{kg\cdot m^2}{s}$ $\displaystyle 7.04*10^{-5}\ \frac{kg\cdot m^2}{s}$

$1.42*10^{-4}\ \frac{kg\cdot m^2}{s}$ $\displaystyle 1.42*10^{-4}\ \frac{kg\cdot m^2}{s}$

$3.872*10^{-4}\ \frac{kg\cdot m^2}{s}$ $\displaystyle 3.872*10^{-4}\ \frac{kg\cdot m^2}{s}$

$3.52*10^{-3}\ \frac{kg\cdot m^2}{s}$ $\displaystyle 3.52*10^{-3}\ \frac{kg\cdot m^2}{s}$

$2.58*10^{-3}\ \frac{kg\cdot m^2}{s}$ $\displaystyle 2.58*10^{-3}\ \frac{kg\cdot m^2}{s}$

Correct answer:

$3.872*10^{-4}\ \frac{kg\cdot m^2}{s}$ $\displaystyle 3.872*10^{-4}\ \frac{kg\cdot m^2}{s}$

Explanation:

The angular momentum of an object is the moment of inertia times the angular velocity:

$L=I\omega$ $\displaystyle L=I\omega$

We can find our moment of inertia using the given formula:

$I_{disc}=\frac{1}{2}MR^2$ $\displaystyle I_{disc}=\frac{1}{2}MR^2$

Plug in the given mass and radius.

$I_{disc}=\frac{1}{2}(0.02kg)(0.11m)^2$ $\displaystyle I_{disc}=\frac{1}{2}(0.02kg)(0.11m)^2$

$I_{disc}=1.21*10^{-4}\ kg\cdot m^2$ $\displaystyle I_{disc}=1.21*10^{-4}\ kg\cdot m^2$

Using the value of the moment of inertia and the given angular velocity, we can calculate the angular momentum.

$L=I\omega$ $\displaystyle L=I\omega$

$L=(1.21*10^{-4}\ kg\cdot m^2)(3.2\frac{rad}{s})$ $\displaystyle L=(1.21*10^{-4}\ kg\cdot m^2)(3.2\frac{rad}{s})$

$L=3.872*10^{-4}\ \frac{kg\cdot m^2}{s}$ $\displaystyle L=3.872*10^{-4}\ \frac{kg\cdot m^2}{s}$

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High School Physics : Circular Motion

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #51 : Circular Motion

Example Question #52 : Circular Motion

Example Question #53 : Circular Motion

Example Question #191 : Motion And Mechanics

Example Question #32 : Using Circular Motion Equations

Example Question #56 : Circular Motion

Example Question #57 : Circular Motion

Example Question #58 : Circular Motion

Example Question #59 : Circular Motion

Example Question #60 : Circular Motion

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