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High School Physics : Motion and Mechanics

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Example Questions

Example Question #181 : Motion And Mechanics

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

Possible Answers:

There is insufficient information to solve

0.5m

0.89m

1.13m

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 #181 : Motion And Mechanics

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

Possible Answers:

17.74N

25.58N

11.43N

0.04N

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 #182 : 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:

$9.06\frac{m}{s}$

$2.16\frac{m}{s}$

$82\frac{m}{s}$

$0.81\frac{m}{s}$

$3.01\frac{m}{s}$

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 #183 : Motion And Mechanics

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:

0.223N

3.682N

4.602N

1.013N

0.178N

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 #42 : Circular Motion

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:

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

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

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

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

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

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 #184 : 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:

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

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

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

1154.2kg*m^2

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

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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Example Question #185 : Motion And Mechanics

What is the angular velocity of an object rotating at $31\frac{rev}{s}$ ?

Possible Answers:

$97.39\frac{rad}{s}$

$9.87\frac{rad}{s}$

$31\frac{rad}{s}$

$194.78\frac{rad}{s}$

$11160\frac{rad}{s}$

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:

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

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Example Question #42 : Circular Motion

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:

0.88J*m

0.11J*m

0.36J*m

There is insufficient information to solve

0.06J*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

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Example Question #182 : Motion And Mechanics

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:

36.50N

4.056N

10.54N

6.67N

21.09N

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

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Example Question #183 : 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:

$15.91\frac{rad}{s}$

$0.19\frac{rad}{s}$

$12.81\frac{rad}{s}$

$49.32\frac{rad}{s}$

$5.13\frac{rad}{s}$

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.

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

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

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High School Physics : Motion and Mechanics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #181 : Motion And Mechanics

Example Question #181 : Motion And Mechanics

Example Question #182 : Motion And Mechanics

Example Question #183 : Motion And Mechanics

Example Question #42 : Circular Motion

Example Question #184 : Motion And Mechanics

Example Question #185 : Motion And Mechanics

Example Question #42 : Circular Motion

Example Question #182 : Motion And Mechanics

Example Question #183 : Motion And Mechanics

All High School Physics Resources

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