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

← Previous 1 2 3 4 5 Next →

Example Question #21 : Circular Motion

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

Now we can plug in and solve:

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

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

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

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

Report an Error

Example Question #21 : Circular Motion

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The formula for centripetal acceleration is:

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

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

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

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

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

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

Report an Error

Example Question #21 : Circular Motion

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The formula for centripetal force is 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.

F_c=ma_c

14.5N=3.3kg*a_c

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

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

Report an Error

Example Question #21 : Circular Motion

If the centripetal force on a 3.3kg object is 14.5N , and the object is 1.9m from the center of the circle, what is its tangential velocity?

Possible Answers:

$1.51\frac{m}{s}$

$2.083\frac{m}{s}$

$8.25\frac{m}{s}$

$2.87\frac{m}{s}$

$4.34\frac{m}{s}$

Correct answer:

$2.87\frac{m}{s}$

Explanation:

The relationship between centripetal acceleration and tangential velocity is:

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

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

The formula for centripetal force is F_c=ma_c .

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

F_c=ma_c

14.5N=3.3kg*a_c

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

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

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.

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

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

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

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

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

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

Report an Error

Example Question #15 : Using Circular Motion Equations

The torque applied to a wrench is 16N*m . If the force applied to the wrench is 40N , how long is the wrench?

Possible Answers:

2.5m

0.4m

24m

56m

Correct answer:

0.4m

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.

16N*m=(40N)r

$\frac{16N*m}{40N}=r$

0.4m=r

Report an Error

Example Question #231 : High School Physics

If the force applied to a 0.21m wrench is 50N , how much torque is generated?

Possible Answers:

50.21N*m

10.5N*m

21N*m

238.1N*m

0.0042N*m

Correct answer:

10.5N*m

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the values for the force and the length. Using these values, we can multiply to find the torque.

$\tau=50N*0.21m$

$\tau=10.5N*m$

Report an Error

Example Question #17 : Using Circular Motion Equations

The torque applied to a wrench is 14N*m . If the length of the wrench is 0.5m , how much force is applied to it?

Possible Answers:

13.5N

28N

32N

14.5N

Correct answer:

28N

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the total torque and the length of the wrench. Given that the pivot point will be at one end of the wrench, and the force will be applied to the other end, this length can be used as our radius.

14N*m=F*0.5m

$\frac{14N*m}{0.5m}=F$

28N=F

Report an Error

Example Question #18 : Using Circular Motion Equations

How much force is required for a 0.3m hammer to produce 4.2N*m of torque?

Possible Answers:

14N

1.26N

3.9N

45N

Correct answer:

14N

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the length of the hammer (radius of the swing) and the torque produced. Using these values, we can solve for the force required.

4.2N*m=F(0.3m)

$\frac{4.2N*m}{0.3m}=F$

14N=F

Report an Error

Example Question #19 : Using Circular Motion Equations

A top spins so that its angular velocity is $\small 31\frac{rad}{s}$ . If the top has a radius of 0.24m , what is the tangential velocity of a point on the edge of the top?

Possible Answers:

$129.17\frac{m}{s}$

$31\frac{m}{s}$

$1.79\frac{m}{s}$

$29.76\frac{m}{s}$

$7.44\frac{m}{s}$

Correct answer:

$7.44\frac{m}{s}$

Explanation:

The relationship between tangential velocity and angular velocity is:

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

We are given the angular velocity and the radius, allowing us to solve for the linear (or tangential) velocity.

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

$31\frac{rad}{s}=\frac{v}{0.24\frac{m}{rad}}$

$0.24\frac{m}{rad}*31\frac{rad}{s}=v$

$7.44\frac{m}{s}=v$

Report an Error

Example Question #20 : Using Circular Motion Equations

A top spins so that the angular velocity is $\small 31\frac{rad}{s}$ . If the top has a radius of 0.24m , what is the centripetal acceleration acting on a point on the edge of the top?

Possible Answers:

$4004.17\frac{m}{s^2}$

$13.28\frac{m}{s^2}$

$230.64\frac{m}{s^2}$

$3.19\frac{m}{s^2}$

There is insufficient information to solve

Correct answer:

$230.64\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is:

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

We know the radius, but we need to find the tangential velocity. The relationship between tangential velocity and angular velocity is:

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

We are given the angular velocity and the radius, allowing us to solve for the linear (or tangential) velocity.

$31\frac{rad}{s}=\frac{v}{0.24\frac{m}{rad}}$

$0.24\frac{m}{rad}*31\frac{rad}{s}=v$

$7.44\frac{m}{s}=v$

Now we have the linear velocity and the radius, allowing us to use the first equation to find the centripetal acceleration.

$a_c=\frac{(7.44\frac{m}{s})^2}{0.24m}$

$a_c=\frac{55.35\frac{m^2}{s^2}}{0.24m}$

$a_c=230.64\frac{m}{s^2}$

Report an Error

← Previous 1 2 3 4 5 Next →

View Tutors

William
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor in Arts, Biological and Physical Sciences.

View Tutors

Karon
Certified Tutor

Florida Atlantic University, Bachelor of Science, Liberal Arts and Sciences. Nova Southeastern University, Master of Science,...

View Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, 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 Atlanta, Computer Science Tutors in San Diego, GMAT Tutors in San Diego, Reading Tutors in Chicago, Statistics Tutors in Atlanta, SSAT Tutors in Seattle, GRE Tutors in Los Angeles, ACT Tutors in Denver, Chemistry Tutors in San Francisco-Bay Area, Statistics Tutors in Miami

Popular Courses & Classes

ACT Courses & Classes in Miami, ISEE Courses & Classes in Miami, Spanish Courses & Classes in Philadelphia, GRE Courses & Classes in Boston, LSAT Courses & Classes in Los Angeles, SSAT Courses & Classes in San Diego, MCAT Courses & Classes in Seattle, SSAT Courses & Classes in Seattle, SSAT Courses & Classes in Phoenix, GRE Courses & Classes in Houston

Popular Test Prep

GRE Test Prep in Houston, GRE Test Prep in Washington DC, SSAT Test Prep in Chicago, LSAT Test Prep in Seattle, SSAT Test Prep in Phoenix, SAT Test Prep in Philadelphia, SAT Test Prep in New York City, SSAT Test Prep in Boston, ISEE Test Prep in Washington DC, GMAT Test Prep in Seattle

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 #21 : Circular Motion

Example Question #21 : Circular Motion

Example Question #21 : Circular Motion

Example Question #21 : Circular Motion

Example Question #15 : Using Circular Motion Equations

Example Question #231 : High School Physics

Example Question #17 : Using Circular Motion Equations

Example Question #18 : Using Circular Motion Equations

Example Question #19 : Using Circular Motion Equations

Example Question #20 : 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: