We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Using Circular Motion Equations

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 #1 : Using Circular Motion Equations

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

Plug in the given values and solve for the acceleration.

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

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

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

Report an Error

Example Question #1 : Using Circular Motion Equations

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

Possible Answers:

18N

$9\pi N$

Correct answer:

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 F_c=ma_c .

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

Plug in the given values and solve for the acceleration.

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

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

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

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

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

F_c=18N

Report an Error

Example Question #222 : High School Physics

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

Possible Answers:

$2\pi s$

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

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

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

$\pi s$

Correct answer:

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

Explanation:

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

Plug in the given values to solve for the period.

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

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

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

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

Report an Error

Example Question #14 : Circular Motion

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

Plug in the given values and solve for the acceleration.

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

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

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

Report an Error

Example Question #16 : Circular Motion

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

Possible Answers:

64N

256N

98N

128N

Correct answer:

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 F_c=ma_c .

Plug in the given values and solve for the acceleration.

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

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

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

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

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

F_c=256N

Report an Error

Example Question #3 : Using Circular Motion Equations

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

Possible Answers:

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

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

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

$16\pi s$

$8\pi s$

Correct answer:

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

Explanation:

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

Plug in the given values to solve for the period.

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

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

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

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

Report an Error

Example Question #162 : Motion And Mechanics

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

Possible Answers:

$6\frac{m}{s}$

$2.45\frac{m}{s}$

$3\frac{m}{s}$

$3.99\frac{m}{s}$

$1.82\frac{m}{s}$

Correct answer:

$2.45\frac{m}{s}$

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 F_c=ma_c .

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

F_c=ma_c

12N=(2kg)a_c

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

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

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

Plug in the given values and solve for the velocity.

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

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

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

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

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

Report an Error

Example Question #163 : Motion And Mechanics

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:

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

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

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

$98101.23\frac{km}{hr}$

$589.22\frac{km}{hr}$

Correct answer:

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

Explanation:

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

Plug in the given values and solve for the velocity.

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

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

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

Report an Error

Example Question #164 : Motion And Mechanics

A large burst of wind exerts 750N of force on a tree 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

75000N*m

7500N*m

75N*m

740N*m

Correct answer:

7500N*m

Explanation:

The formula for torque is:

$\tau=F*r$

Plug in our given values:

$\tau=750N*10m$

$\tau=7500N*m$

Report an Error

Example Question #4 : Using Circular Motion Equations

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

Possible Answers:

34.78N*m

1.84N*m

0.92N*m

8.23N*m

7.77N*m

Correct answer:

1.84N*m

Explanation:

The formula for torque is:

$\tau=F*r$

Plug in our given values:

$\tau=8N*0.23m$

$\tau=1.84N*m$

Report an Error

View Tutors

Oselen
Certified Tutor

UNIVERSITY OF BENIN, Bachelor of Science, Mechanical Engineering. University of Windsor, Master of Science, Mechanical Engine...

View Tutors

George
Certified Tutor

Pensacola State College, Bachelor in Arts, Mathematics.

View Tutors

Krittika
Certified Tutor

University of Dhaka, Master's/Graduate, Statistics. University of Waterloo, Master's/Graduate, Biostatistics.

All High School Physics Resources

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

Popular Subjects

Math Tutors in Phoenix, Chemistry Tutors in Dallas Fort Worth, Computer Science Tutors in Boston, Chemistry Tutors in San Francisco-Bay Area, Reading Tutors in Denver, Reading Tutors in Seattle, Biology Tutors in Denver, Algebra Tutors in San Francisco-Bay Area, SAT Tutors in Seattle, Algebra Tutors in Miami

Popular Courses & Classes

LSAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Boston, GRE Courses & Classes in Seattle, SAT Courses & Classes in Denver, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in Washington DC, GMAT Courses & Classes in Phoenix, SSAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in New York City, SAT Courses & Classes in Atlanta

Popular Test Prep

ACT Test Prep in Miami, GMAT Test Prep in Phoenix, ISEE Test Prep in Washington DC, SSAT Test Prep in Philadelphia, ACT Test Prep in Philadelphia, ACT Test Prep in Washington DC, LSAT Test Prep in Miami, ISEE Test Prep in Atlanta, GMAT Test Prep in Miami, GMAT Test Prep in Dallas Fort Worth

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 : Using Circular Motion Equations

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Using Circular Motion Equations

Example Question #1 : Using Circular Motion Equations

Example Question #222 : High School Physics

Example Question #14 : Circular Motion

Example Question #16 : Circular Motion

Example Question #3 : Using Circular Motion Equations

Example Question #162 : Motion And Mechanics

Example Question #163 : Motion And Mechanics

Example Question #164 : Motion And Mechanics

Example Question #4 : Using Circular Motion Equations

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: