We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Circular Motion

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Circular Motion

If 11N*m of torque is generated using a 0.43m wrench, how much force is required?

Possible Answers:

0.04N

17.74N

25.58N

11.43N

12.79N

Correct answer:

25.58N

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the length of the lever arm and the torque generated, allowing us to solve for the force applied.

11N*m=F(0.43m)

$\frac{11N*m}{0.43m}=F$

25.58N

Report an Error

Example Question #22 : Using Circular Motion Equations

A 1200kg car makes a right turn. The radius of this curve is 4.2m . If the force of friction between the tires and the road is 2587.2N , what is the maximum velocity that the car can have before skidding?

Possible Answers:

$3.01\frac{m}{s}$

$0.81\frac{m}{s}$

$2.16\frac{m}{s}$

$82\frac{m}{s}$

$9.06\frac{m}{s}$

Correct answer:

$3.01\frac{m}{s}$

Explanation:

To solve this problem, recognize that the force due to friction must equal the centripetal force of the curve:

F_f=F_c

This will give the maximum force that the car can have in the curve without skidding. Expand the equation for centripetal force.

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

We are given the value for the force of friction, the mass of the car, and the radius of the curve. Using these values, we can find the velocity.

$2587.2N=1200kg*\frac{v^2}{4.2m}$

$\frac{2587.2N}{1200kg}=\frac{v^2}{4.2m}$

$2.156\frac{m}{s^2}=\frac{v^2}{4.2m}$

$4.2m*2.156\frac{m}{s^2}=v^2$

$9.0552\frac{m^2}{s^2}=v^2$

$\sqrt{9.0552\frac{m^2}{s^2}}=\sqrt{v^2}$

$3.01\frac{m}{s}=v$

Report an Error

Example Question #21 : Using Circular Motion Equations

A 0.22kg rock is swung in a circle on a 0.8m long rope. How much centripetal force is required for the rock to maintain a velocity of $0.9\frac{m}{s}$ ?

Possible Answers:

1.013N

0.223N

3.682N

0.178N

4.602N

Correct answer:

0.223N

Explanation:

The equation for centripetal force is:

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

We are given the radius (length of the rope), velocity, and mass of the rock, allowing us to calculate the centripetal force.

$F_c=0.22kg*\frac{(0.9\frac{m}{s})^2}{0.8m}$

$F_c=0.22kg*\frac{0.81\frac{m^2}{s^2}}{0.8m}$

$F_c=0.22kg*1.0125\frac{m}{s^2}$

F_c=0.223N

Report an Error

Example Question #184 : Motion And Mechanics

A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the angular velocity of the satellite's orbit?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Possible Answers:

$6.54*10^{-14}\frac{rad}{s}$

$3.68*10^{-10}\frac{rad}{s}$

$1.45*10^{17}\frac{rad}{s}$

$1.44*10^{-10}\frac{rad}{s}$

$5.33*10^7\frac{rad}{s}$

Correct answer:

$3.68*10^{-10}\frac{rad}{s}$

Explanation:

Angular velocity is given by the equation:

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

We know the radius, but we need to find the tangential velocity. We can do this be finding the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means F_c=F_G .

Solve for F_G for the satellite. To do this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

Now that we know the force, we can find the acceleration. Remember that centripetal force is F_c=m*a_c . Set our two forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

Now we can find the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

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

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

Now that we know the tangential velocity, we can divide by the radius to find the angular velocity. Again, remember that the radius of the orbit is equal to the sum of the Earth's radius and the height of the satellite above the surface.

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

$\omega=\frac{5.33*10^7\frac{m}{s}}{1.45*10^{17}m}$

$\omega=3.68*10^{-10}\frac{rad}{s}$

Report an Error

Example Question #185 : Motion And Mechanics

A 1.5kg baseball has a radius of 0.038m . What is the moment of inertia for the ball?

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

Possible Answers:

$2.3*10^{-2}kg*m^2$

$5.7*10^{-2}kg*m^2$

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

$4.3*10^{-4}kg*m^2$

1154.2kg*m^2

Correct answer:

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

Explanation:

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.

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

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

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

Report an Error

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:

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

Report an Error

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

Report an Error

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

Report an Error

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.

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

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

Report an Error

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.

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

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

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

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

Report an Error

View Tutors

Dereika
Certified Tutor

Southern University and A & M College, Bachelor of Science, Accounting. Grand Canyon University, Master of Science, Accounting.

View Tutors

Alexis
Certified Tutor

CUNY Lehman College, Bachelor in Arts, Linguistics.

View Tutors

Dejuan
Certified Tutor

University of Illinois at Urbana-Champaign, Bachelor of Science, Electrical Engineering.

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Physics Tutors in Seattle, Calculus Tutors in Chicago, SSAT Tutors in Denver, French Tutors in Boston, Reading Tutors in Denver, Statistics Tutors in Denver, Chemistry Tutors in Dallas Fort Worth, Calculus Tutors in San Francisco-Bay Area, Statistics Tutors in Philadelphia, Computer Science Tutors in San Diego

Popular Courses & Classes

Spanish Courses & Classes in Chicago, ISEE Courses & Classes in Miami, MCAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in San Diego, ACT Courses & Classes in Los Angeles, GMAT Courses & Classes in Washington DC, ISEE Courses & Classes in Denver, MCAT Courses & Classes in Phoenix, MCAT Courses & Classes in Boston, SAT Courses & Classes in Houston

Popular Test Prep

ACT Test Prep in Boston, LSAT Test Prep in Philadelphia, ACT Test Prep in Los Angeles, GRE Test Prep in Seattle, GRE Test Prep in Boston, GMAT Test Prep in Seattle, GMAT Test Prep in Boston, SAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Houston, GMAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Physics : Circular Motion

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #41 : Circular Motion

Example Question #22 : Using Circular Motion Equations

Example Question #21 : Using Circular Motion Equations

Example Question #184 : Motion And Mechanics

Example Question #185 : Motion And Mechanics

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

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: