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High School Physics : Using Circular Motion Equations

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Example Questions

Example Question #166 : Motion And Mechanics

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:

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

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

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

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

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

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:

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

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

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

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

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Example Question #161 : Motion And Mechanics

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:

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

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

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

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

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

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.

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

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

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

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

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Example Question #168 : Motion And Mechanics

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:

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

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

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

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

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

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

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

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

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Example Question #162 : Motion And Mechanics

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:

$8.25\frac{m}{s}$

$1.51\frac{m}{s}$

$2.083\frac{m}{s}$

$4.34\frac{m}{s}$

$2.87\frac{m}{s}$

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.

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

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

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

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

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

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

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Example Question #171 : Motion And Mechanics

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

24m

0.4m

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

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Example Question #172 : Motion And Mechanics

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

Possible Answers:

10.5N*m

238.1N*m

50.21N*m

21N*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.

$\tau=50N*0.21m$

$\tau=10.5N*m$

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Example Question #11 : 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

32N

28N

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

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Example Question #174 : Motion And Mechanics

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

Possible Answers:

14N

3.9N

1.26N

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

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Example Question #31 : Circular Motion

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:

$129.17\frac{m}{s}$

$7.44\frac{m}{s}$

$29.76\frac{m}{s}$

$1.79\frac{m}{s}$

$31\frac{m}{s}$

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.

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

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

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

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

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Example Question #35 : Circular Motion

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:

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

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

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

There is insufficient information to solve

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

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.

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

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

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

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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 #166 : Motion And Mechanics

Example Question #161 : Motion And Mechanics

Example Question #168 : Motion And Mechanics

Example Question #162 : Motion And Mechanics

Example Question #171 : Motion And Mechanics

Example Question #172 : Motion And Mechanics

Example Question #11 : Using Circular Motion Equations

Example Question #174 : Motion And Mechanics

Example Question #31 : Circular Motion

Example Question #35 : Circular Motion

All High School Physics Resources

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