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High School Physics : Using Circular Motion Equations

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Example Question #31 : 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 force acting on a point on the edge of the top?

Possible Answers:

961N

55.35N

230.64N

There is insufficient information to solve

3.19N

Correct answer:

There is insufficient information to solve

Explanation:

The formula for centripetal force is:

$F_c=ma_c=\frac{mv^2}{r}$

While we can find what the acceleration will be, there is no way for us to find the mass. The force cannot be calculated.

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Example Question #32 : Circular Motion

A girl sits on the edge of a merry-go-round. If she travels at $0.02\frac{m}{s}$ and the radius of her motion is 2.1m , what is the centripetal acceleration acting upon her?

Possible Answers:

$5.26\frac{m}{s^2}$

$0.019\frac{m}{s^2}$

$0.0004\frac{m}{s^2}$

$0.00036\frac{m}{s^2}$

$0.00019\frac{m}{s^2}$

Correct answer:

$0.00019\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is:

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

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

$a_c=\frac{(0.02\frac{m}{s})^2}{2.1m}$

$a_c=\frac{0.0004\frac{m^2}{s^2}}{2.1m}$

$a_c=0.00019\frac{m}{s^2}$

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Example Question #33 : Circular Motion

A 41kg girl sits on the edge of a merry-go-round. If she travels at $0.02\frac{m}{s}$ and the radius of her motion is 2.1m , what is the centripetal force acting upon her?

Possible Answers:

0.0078N

1.244N

0.016N

210.5N

0.661N

Correct answer:

0.0078N

Explanation:

The formula for centripetal force is:

$F_c=\frac{mv^2}{r}$

We are given the girl's mass, her linear velocity, and the radius of her motion. Using these values, we can calculate the centripetal force.

$F_c=\frac{(41kg)(0.02\frac{m}{s})^2}{2.1m}$

$F_c=\frac{(41kg)(0.0004\frac{m^2}{s^2})}{2.1m}$

F_c=0.0078N

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Example Question #171 : Motion And Mechanics

How much torque is generated by applying 13.2N of force to the end of a 1.1m lever?

Possible Answers:

14.52N*m

12.1N*m

13.2N*m

15.97N*m

12N*m

Correct answer:

14.52N*m

Explanation:

The equation for torque is:

$\tau=F*r$

We are given the force and distance (radius), allowing us to solve for the torque by multiplication.

$\tau=13.2N*1.1m$

$\tau=14.52N*m$

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Example Question #31 : Circular Motion

If 16N*m of torque is generated by applying 18N of force, how long is the lever arm?

Possible Answers:

1.13m

0.89m

There is insufficient information to solve

0.5m

Correct answer:

0.89m

Explanation:

The formula for torque is:

$\tau=F*r$

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

16N*m=(18N)r

$\frac{16N*m}{18N}=r$

0.89m=r

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Example Question #41 : Circular Motion

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

Possible Answers:

17.74N

11.43N

0.04N

25.58N

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

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Example Question #181 : Motion And Mechanics

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:

$3.01\frac{m}{s}$

$82\frac{m}{s}$

$2.16\frac{m}{s}$

$9.06\frac{m}{s}$

$0.81\frac{m}{s}$

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.

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

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

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

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

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

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

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

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Example Question #23 : 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:

0.178N

3.682N

0.223N

1.013N

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

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Example Question #21 : Using Circular Motion Equations

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:

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

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

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

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

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

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.

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

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

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

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Example Question #25 : Using Circular Motion Equations

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:

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

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

1154.2kg*m^2

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

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

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.

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

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

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

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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 #31 : Circular Motion

Example Question #32 : Circular Motion

Example Question #33 : Circular Motion

Example Question #171 : Motion And Mechanics

Example Question #31 : Circular Motion

Example Question #41 : Circular Motion

Example Question #181 : Motion And Mechanics

Example Question #23 : Using Circular Motion Equations

Example Question #21 : Using Circular Motion Equations

Example Question #25 : Using Circular Motion Equations

All High School Physics Resources

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