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High School Physics : Using Circular Motion Equations

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Example Questions

Example Question #42 : Circular Motion

What is the angular velocity of an object rotating at $31\frac{rev}{s}$ ?

Possible Answers:

$9.87\frac{rad}{s}$

$31\frac{rad}{s}$

$97.39\frac{rad}{s}$

$194.78\frac{rad}{s}$

$11160\frac{rad}{s}$

Correct answer:

$194.78\frac{rad}{s}$

Explanation:

The problem gives us a value in terms of revolutions per second, but we need radians per second.

Remember that $1rev=2\pi rad$ so we can convert:

$31\frac{rev}{s}*\frac{2\pi rad}{1rev}=62\pi\frac{rad}{s}=194.78\frac{rad}{s}$

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Example Question #186 : Motion And Mechanics

A 1.5kg baseball has a radius of 0.038m . If it spins at a rate of $65\frac{rev}{s}$ , what is its angular momentum?

$I_{sphere}=\frac{2}{5}mr^2$

Possible Answers:

There is insufficient information to solve

0.88J*m

0.06J*m

0.11J*m

0.36J*m

Correct answer:

0.36J*m

Explanation:

The formula for angular momentum is:

$L=I\omega$

We need to find values for the moment of inertia and the angular velocity in order to find the angular momentum.

The given equation for moment of inertia is:

$I_{sphere}=\frac{2}{5}mr^2$

Use the given values for the mass and radius of the ball to solve for the moment of inertia.

$I=\frac{2}{5}mr^2$

$I=\frac{2}{5}(1.5kg)(0.038m)^2$

$I=8.7*10^{-4}kg*m^2$

Now we need to find the angular velocity. The problem gives us a value in terms of revolutions per second, but we need radians per second.

Remember that $1rev=2\pi rad$ so we can convert:

$65\frac{rev}{s}*\frac{2\pi rad}{1rev}=130\pi\frac{rad}{s}=408.4\frac{rad}{s}$

Now we can use our values for angular velocity and moment of inertia in the equation for angular momentum.

$L=I\omega$

$L=(8.7*10^{-4}kg*m^2)(408.4\frac{rad}{s})$

L=0.36J*m

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Example Question #43 : Circular Motion

A ball of mass $\small 2.34kg$ is tied to a massless string of length $\small 3m$ and swings in a circular motion moving at $\small 5.2\frac{m}{s}$ . What is the tension in the string?

Possible Answers:

4.056N

21.09N

36.50N

10.54N

6.67N

Correct answer:

21.09N

Explanation:

The tension in the string will be equal to the centripetal force on the ball.

F_c=F_T

Centripetal force is given by the formula:

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

We are given the mass of the ball, its velocity, and the length of the string, which will determine the radius.

$F_c=(2.34kg)\frac{(5.2\frac{m}{s})^2}{3m}$

$F_c=(2.34kg)(9.01\frac{m}{s^2})$

F_c=21.09N

Since the centripetal force and force of tension will be equal, this is our answer.

F_c=F_T=21.09N

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Example Question #188 : Motion And Mechanics

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

What is the angular velocity of the object?

Possible Answers:

$5.13\frac{rad}{s}$

$12.81\frac{rad}{s}$

$0.19\frac{rad}{s}$

$15.91\frac{rad}{s}$

$49.32\frac{rad}{s}$

Correct answer:

$5.13\frac{rad}{s}$

Explanation:

The relationship between linear and angular velocity is:

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

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

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

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

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

Plug that into our initial equation, along with the radius, to convert it to angular velocity.

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

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

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Example Question #44 : Circular Motion

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

What is the centripetal acceleration on the object?

Possible Answers:

$49.32\frac{m}{s^2}$

$5.13\frac{m}{s^2}$

$81.65\frac{m}{s^2}$

$253.13\frac{m}{s^2}$

$26.34\frac{m}{s^2}$

Correct answer:

$81.65\frac{m}{s^2}$

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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Example Question #51 : Circular Motion

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

What is the angular momentum of the object?

Possible Answers:

$179.63\,J*s$

$11.29\,J*s$

$108.5\,J*s$

$57.9\,J*s$

$5.13\,J*s$

Correct answer:

$108.5\,J*s$

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

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

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

$L=108.5\,J*s$

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Example Question #193 : Motion And Mechanics

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

What is the moment of inertia on this object?

Possible Answers:

6.82kg*m^2

15kg*m^2

21.14kg*m^2

5.13kg*m^2

9.61kg*m^2

Correct answer:

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

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

I=(2.2kg)(3.1m)^2

I=(2.2kg)(9.61m^2)

I=21.14kg*m^2

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Example Question #194 : Motion And Mechanics

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

What is the period of the object's orbit?

Possible Answers:

1.5s

0.61s

0.82s

0.39s

1.22s

Correct answer:

1.22s

Explanation:

The relationship between period and angular velocity is:

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

The relationship between linear and angular velocity is:

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

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

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

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

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

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

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

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

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

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

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

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

T=1.22s

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Example Question #195 : Motion And Mechanics

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

What is the angular displacement of the object after ?

Possible Answers:

79.55rad

161.16rad

4.08rad

26.65rad

9234rad

Correct answer:

26.65rad

Explanation:

Angular displacement is equal to the angular velocity times time:

$\theta=\omega*\Delta t$

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

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

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

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

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

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

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

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

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

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

$\theta=\omega *\Delta t$

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

$\theta=26.65rad$

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Example Question #196 : Motion And Mechanics

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

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

Possible Answers:

$11.48\, rev$

$17.08\, rev$

$1.22\, rev$

$0.09\, rev$

$72.1 \, rev$

Correct answer:

$11.48\, rev$

Explanation:

We can set up a proportion here:

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

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

The relationship between period and angular velocity is:

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

The relationship between linear and angular velocity is:

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

Cross multiply.

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

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

$11.48\, rev = x$

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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 #42 : Circular Motion

Example Question #186 : Motion And Mechanics

Example Question #43 : Circular Motion

Example Question #188 : Motion And Mechanics

Example Question #44 : Circular Motion

Example Question #51 : Circular Motion

Example Question #193 : Motion And Mechanics

Example Question #194 : Motion And Mechanics

Example Question #195 : Motion And Mechanics

Example Question #196 : Motion And Mechanics

All High School Physics Resources

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