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High School Physics : Motion and Mechanics

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Example Questions

Example Question #191 : 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 centripetal acceleration on the object?

Possible Answers:

$26.34\frac{m}{s^2}$

$81.65\frac{m}{s^2}$

$5.13\frac{m}{s^2}$

$253.13\frac{m}{s^2}$

$49.32\frac{m}{s^2}$

Correct answer:

$81.65\frac{m}{s^2}$

Explanation:

Centripetal acceleration is equal to the tangential velocity squared over the radius:

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

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

$35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$

Use the linear velocity and the radius in the initial equation to solve for the centripetal acceleration.

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

$a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$

$a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$

$a_c=81.65\frac{m}{s^2}$

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Example Question #192 : 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 momentum of the object?

Possible Answers:

$11.29\,J*s$

$57.9\,J*s$

$108.5\,J*s$

$179.63\,J*s$

$5.13\,J*s$

Correct answer:

$108.5\,J*s$

Explanation:

There is a direct relationship between angular momentum and linear momentum. Angular momentum is equal to the linear momentum times the radius:

L=p*r

We are given the value of both the linear momentum and the radius, allowing us to solve for the angular momentum.

$L=(35kg*\frac{m}{s})(3.1m)$

$L=108.5\,J*s$

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Example Question #193 : 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 moment of inertia on this object?

Possible Answers:

5.13kg*m^2

15kg*m^2

21.14kg*m^2

9.61kg*m^2

6.82kg*m^2

Correct answer:

21.14kg*m^2

Explanation:

Since the object is moving in a perfect circle and not rotating about some fixed point within itself (like spinning a ball or a frisbee), the equation for moment of inertia is:

I=mr^2

We are given the mass and radius, allowing us to calculate the moment of inertia from this equation.

I=(2.2kg)(3.1m)^2

I=(2.2kg)(9.61m^2)

I=21.14kg*m^2

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Example Question #194 : 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 period of the object's orbit?

Possible Answers:

0.39s

0.61s

1.5s

1.22s

0.82s

Correct answer:

1.22s

Explanation:

The relationship between period and angular velocity is:

$\omega=\frac{2\pi}{T}$

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 the equation for angular velocity, along with the radius.

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

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

Plug this term back into the initial equation to solve for the period.

$\omega=\frac{2\pi}{T}$

$5.13\frac{rad}{s}=\frac{2\pi}{T}$

$T=\frac{2\pi}{5.13\frac{rad}{s}}$

T=1.22s

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Example Question #51 : Circular Motion

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 displacement of the object after ?

Possible Answers:

161.16rad

4.08rad

79.55rad

9234rad

26.65rad

Correct answer:

26.65rad

Explanation:

Angular displacement is equal to the angular velocity times time:

$\theta=\omega*\Delta t$

We know the time, but we need to solve for the angular velocity. 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 the equation for angular velocity, along with the radius.

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

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

Now that we have the angular velocity, we can solve for the angular displacement at the given time.

$\theta=\omega *\Delta t$

$\theta=5.13\frac{rad}{s}*5s$

$\theta=26.65rad$

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Example Question #195 : 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 $35\frac{kg*m}{s}$ .

How many revolutions does the object go through in 14s ?

Possible Answers:

$0.09\, rev$

$11.48\, rev$

$17.08\, rev$

$1.22\, rev$

$72.1 \, rev$

Correct answer:

$11.48\, rev$

Explanation:

We can set up a proportion here:

$\frac{1rev}{T}=\frac{x\, rev}{14s}$ .

In other words, the object can do one revolution per period (), so it can do revolutions in 14s . First we need to find the period.

The relationship between period and angular velocity is:

$\omega=\frac{2\pi}{T}$

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 the equation for angular velocity, along with the radius.

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

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

Plug this term back into the first equation to solve for the period.

$\omega=\frac{2\pi}{T}$

$5.13\frac{rad}{s}=\frac{2\pi}{T}$

$T=\frac{2\pi}{5.13\frac{rad}{s}}$

T=1.22s

Now that we know the period, we can return to the proportion to solve for the rotations completed in the given time.

$\frac{1rev}{1.22s}=\frac{x\, rev}{14s}$

Cross multiply.

$14\, rev*s=(1.22s)(x)$

$\frac{14\,rev*s}{1.22s}=x$

$11.48\, rev = x$

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Example Question #41 : Using Circular Motion Equations

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 centripetal force on the object?

Possible Answers:

129.36N

179.63N

81.65N

15.91N

395.19N

Correct answer:

179.63N

Explanation:

Newton's second law states that F=ma . We know the mass of the object, but we need to find the centripetal acceleration to calculate the centripetal force.

Centripetal acceleration is equal to the tangential velocity squared over the radius:

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

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

$35\frac{kg*m}{s}=(2.2kg)v$

$\frac{35\frac{kg*m}{s}}{2.2kg}=v$

$15.91\frac{m}{s}=v$

Use the linear velocity and the radius in the previous equation to solve for the centripetal acceleration.

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

$a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$

$a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$

$a_c=81.65\frac{m}{s^2}$

Use the centripetal acceleration in Newton's second law, along with the mass, to calculate the centripetal force.

F_c=ma_c

$F_c=2.2kg*81.65\frac{m}{s^2}$

F_c=179.63N

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Example Question #195 : Motion And Mechanics

A baseball has an angular velocity of $88\frac{rad}{s}$ . If the baseball reaches the plate 0.45s after the pitcher releases it, what is the angular displacement of the ball?

Possible Answers:

$44\: rad$

$39.6\: rad$

$2.83\: rad$

$248.8\: rad$

Correct answer:

$39.6\: rad$

Explanation:

The formula for angular displacement with no angular acceleration is $\theta=\omega*t$ .

Plug in our given values.

$\theta=\omega*t$

$\theta=88\frac{rads}{s}*0.45s$

$\theta =39.6\: rad$

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Example Question #43 : Using Circular Motion Equations

A baseball batter swings and misses at a baseball. If the length from his shoulder to the end of the bat is 1.6m , the bat has a mass of 1.1kg , and the bat moves with a linear velocity of $18.3\frac{m}{s}$ , what is the angular momentum generated?

Possible Answers:

32.21J*s

0.03J*s

16.54J*s

21.0J*s

10.39J*s

Correct answer:

32.21J*s

Explanation:

Angular momentum is the product of linear momentum times the lever arm:

L=p*r

Expand this equation by using the formula for linear momentum.

p=mv

L=mv*r

We are given the length of the lever arm (radius), the mass of the bat, and the linear velocity of the swing. Using these values, we can solve for the angular momentum.

L=mv*r

$L=(1.1kg)(18.3\frac{m}{s})(1.6m)$

L=32.21J*s

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Example Question #42 : Using Circular Motion Equations

A child swings a pail of water in a circle, using her shoulder as the pivot point. If the child's arm is 0.8m long and the pail of water has a linear momentum of $1.67\frac{kg\cdot m}{s}$ , what is the angular momentum of the pail of water?

Possible Answers:

$2.1\ \frac{kg\cdot m^2}{s}$

$1.3\ \frac{kg\cdot m^2}{s}$

$1.07\ \frac{kg\cdot m^2}{s}$

$0.87\ \frac{kg\cdot m^2}{s}$

$2.23\ \frac{kg\cdot m^2}{s}$

Correct answer:

$1.3\ \frac{kg\cdot m^2}{s}$

Explanation:

Angular momentum is equal to the linear momentum times the radial arm.

L=r*p

Since the pivot point of the child is the shoulder, the radial arm is the child's arm length. We are given this length, as well as the linear momentum, allowing us to solve for the angular momentum.

$L=(0.8m)(1.67\frac{kg\cdot m}{s})$

$L=1.3\frac{kg\cdot m^2}{s}$

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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 #191 : Motion And Mechanics

Example Question #192 : Motion And Mechanics

Example Question #193 : Motion And Mechanics

Example Question #194 : Motion And Mechanics

Example Question #51 : Circular Motion

Example Question #195 : Motion And Mechanics

Example Question #41 : Using Circular Motion Equations

Example Question #195 : Motion And Mechanics

Example Question #43 : Using Circular Motion Equations

Example Question #42 : Using Circular Motion Equations

All High School Physics Resources

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