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AP Physics 1 : Newtonian Mechanics

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Example Questions

Example Question #1041 : Ap Physics 1

Vertical circle

A ball of mass $m=0.22\;kg$ is being swung in a vertical circle on a string, as shown in the given figure. At the instant it is directly above the center of the circle it is making, its speed is $v=2.7\frac{m}{s}$ . The string has negligible mass, and its length is $\l=0.61\,m$ . What is the tension, , in the string at the instant the ball is at the top of its circle?

Possible Answers:

$T=2.63\,N$

$T=0\,N$

$T=4.79\,N$

$T=2.16\,N$

$T=0.47\,N$

Correct answer:

$T=0.47\,N$

Explanation:

The tension and gravity both point down toward the center of the circle. The net force on the ball is the vector sum of these two:

$\Sigma F=T+mg$

For an object in circular motion, the net force must be equal to:

$\Sigma F=\frac{mv^2}{r}$

In this case, the force of gravity is helping the tension provide this force, so the tension is not as great as it would be if it were supplying the entire centripetal force.

$\frac{mv^2}{r}=T+mg$

Substitute and solve.

$T=\frac{mv^2}{r}-mg=\frac{0.22\times 2.1^2}{.61}-0.22\times9.8=0.47\,N$

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Example Question #51 : Circular And Rotational Motion

When a road is dry, a particular car can safely navigate a turn with a 50 m radius of curvature at $20 \frac{m}{s}$ without slipping. What is the coefficient of friction if this is the fastest speed the car can take this turn? Is this the static or kinetic coefficient?

Possible Answers:

0.04 , static

0.82 , static

0.82 , kinetic

0.04 , kinetic

0.5 , kinetic

Correct answer:

0.82 , static

Explanation:

For the car going around a flat turn, the force of friction is the only force keeping the car in a circular path. This lets us equate the centripetal force to the force of friction.

F_c=F_f

$\frac{mv^2}{r}=\mu N$

N=mg

Substitute and solve for the coefficient of friction.

$\mu=\frac{v^2}{gr}=\frac{(20\frac{m}{s})^2}{9.81\frac{m}{s^2}(50m)}=0.82$

Since the tires are rotating but not slipping on the surface, this is the coefficient of static friction, $\mu_s$ .

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Example Question #21 : Centripetal Force And Acceleration

On a wet road, the coefficient of static friction between a car's tires and the flat road is $\mu_s=0.4$ . What is the maximum speed a car can safely navigate a turn with a 30m radius of curvature?

Possible Answers:

$3.46 \frac{m}{s}$

$10.85 \frac{m}{s}$

$12.0 \frac{m}{s}$

$26.0 \frac{m}{s}$

$117.7 \frac{m}{s}$

Correct answer:

$10.85 \frac{m}{s}$

Explanation:

When a car is going around a flat turn, the only force keeping the car going along a circular path is the force of friction, which is equal to the centripetal force. We do not need the mass of the car to calculate the maximum speed, as we will show below.

F_c=F_f

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

N=mg

Rearrange these equations and solve for the velocity.

$v = \sqrt{\mu_s g r}$

$v= \sqrt{0.4(9.81\frac{m}{s^2})30m}=10.85\frac{m}{s}$

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Example Question #22 : Centripetal Force And Acceleration

If a car can safely go around an icy banked $\left(30^{\circ} \right )$ turn at a speed of $24 \frac{m}{s}$ , what is the radius of curvature $\left(r \right )$ of the turn?

Possible Answers:

67.8m

Cannot be determined without knowing the mass of the car

117.4m

4.24m

102m

Correct answer:

102m

Explanation:

For a car going around a banked turn with no friction, a component of the normal force is what provides the centripetal force. The radius of the curve can be calculated without knowing the mass of the car.

$mg\tan{\theta}=m\frac{v^2}{r}$

$r=\frac{v^2}{g\tan{\theta}}$

$r=\frac{\left(24\frac{m}{s} \right )^2}{9.81\frac{m}{s^2}\tan{30^{\circ}}}=102m$

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Example Question #54 : Circular And Rotational Motion

You tie a 1.2 kg rock to a rope and swing it above your head with a frequency of $\frac{1.1rotations}{ second}$ . The distance from your hand to the rope is .45 meters . Determine the centripital force experience by the rock.

Possible Answers:

9.1 N

31.3 N

12.2 N

17.6 N

None of these

Correct answer:

17.6 N

Explanation:

Centripital acceleration is

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

Force is defined as

F=ma

Using subsitution

$F=m\frac{v^2}{r}$

The rock has a frequency of $\frac{1.1rotations}{ second}$ .

Each circle it makes has a circumference of $2\cdot\pi\cdot .45meters$ , which is equivalent to 2.82 meters .

That means it travel 2.82 meters in 1.1 seconds , which is $2.57 \frac{meters}{second}$

Plugging in our values:

$F=1.2\frac{2.57^2}{.45}$

F=17.6 N

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Example Question #23 : Centripetal Force And Acceleration

A ball attached to a 1 meter-long string is being spun around and is traveling at $4\frac{m}{s}$ .

What is the centripetal acceleration of the ball?

Possible Answers:

$\frac{1}{16}\frac{m}{s^2}$

$8\frac{m}{s^2}$

$16\frac{m}{s^2}$

$4\frac{m}{s^2}$

$\frac{1}{4}\frac{m}{s^2}$

Correct answer:

$16\frac{m}{s^2}$

Explanation:

We use the formula for centripetal acceleration:

$a_{c}=\frac{v^{2}}{r}$

Where a_c is the centripetal acceleration, is the velocity $\left(4\frac{m}{s}\right)$ and is the radius (1m).

Plug in these known values and solve.

$a_c=\frac{(4^2\frac{m}{s})}{1m}$

$a_c=16\frac{m}{s^2}$

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Example Question #102 : Circular, Rotational, And Harmonic Motion

A person of mass 95kg is riding a ferris wheel of radius 10m . The wheel is spinning at a constant angular velocity of 1rpm . Determine the force exerted on the rider by their seat at the top of the ferris wheel.

Possible Answers:

$761N=F_{seat}$

$921N=F_{seat}$

None of these

$665N=F_{seat}$

$1001N=F_{seat}$

Correct answer:

$921N=F_{seat}$

Explanation:

Convert $\frac{rotations}{minute}$ to $\frac{meters}{second}$ :

$1\frac{rotation}{minute}*\frac{1 minute}{60 seconds}*\frac{2*pi*10meters}{1 rotation}=\frac{1.05 meters}{second}$

Definition of centripetal acceleration:

$a_{cent}=\frac{v^2}{r}$

Plug in values:

$a_{cent}=\frac{1.05^2}{10}$

$a_{cent}=.11\frac{meters}{second^2}$

Superposition of forces:

$F_{total}=F_1+F_2+...$

$m*a_{cent}=mg+F_{seat}$

Where both $a_{cent}$ and are pointing down, and thus negative.

Solve for $F_{seat}$

$m*a_{cent}-mg=F_{seat}$

Plug in values:

$95*(-.11)-95(-9.8)=F_{seat}$

$921N=F_{seat}$

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Example Question #103 : Circular, Rotational, And Harmonic Motion

Possible Answers:

$844N=F_{seat}$

$887N=F_{seat}$

$1021N=F_{seat}$

$1411N=F_{seat}$

$941N=F_{seat}$

Correct answer:

$941N=F_{seat}$

Explanation:

Convert $\frac{rotations}{minute}$ to $\frac{meters}{second}$ :

$1\frac{rotation}{minute}*\frac{1 minute}{60 seconds}*\frac{2*pi*10meters}{1 rotation}=\frac{1.05 meters}{second}$

Definition of centripetal acceleration:

$a_{cent}=\frac{v^2}{r}$

Plug in values:

$a_{cent}=\frac{1.05^2}{10}$

$a_{cent}=.11\frac{meters}{second^2}$

Superposition of forces:

$F_{total}=F_1+F_2+...$

$m*a_{cent}=mg+F_{seat}$

Where is pointing down, and thus negative. However, $a_{cent}$ is pointing up, and thus positive.

Solve for $F_{seat}$

$m*a_{cent}-mg=F_{seat}$

Plug in values:

$95*(.11)-95(-9.8)=F_{seat}$

$941N=F_{seat}$

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Example Question #24 : Centripetal Force And Acceleration

Radius of earth: 6.37*10^6m

Estimate a centripetal acceleration of someone standing on the equator with respect to the center of the earth.

Possible Answers:

$9.7*10^{-4}\frac{m}{s^2}$

$8.8*10^{-4}\frac{m}{s^2}$

$6.2*10^{-4}\frac{m}{s^2}$

$5.1*10^{-4}\frac{m}{s^2}$

None of these

Correct answer:

$6.2*10^{-4}\frac{m}{s^2}$

Explanation:

A person on the equator travels in a circle with the same radius as the earth's every 24hr .

Thus, distance divided by time:

$v=\frac{2*\pi*r}{t}$

Convert to and plug in values:

$v=\frac{2*\pi*6.37*10^6}{24*3600}$

$v=463\frac{m}{s^2}$

Use the definition of centripetal acceleration:

$a_{cent}=\frac{v^2}{r}$

Plug in values:

$a_{cent}=\frac{463^2}{6.37*10^6}$

$a_{cent}=6.2*10^{-4}\frac{m}{s^2}$

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Example Question #21 : Centripetal Force And Acceleration

Earth mass: $5.97*10^{24}kg$

Earth distance from the Sun: 149.6*10^9m

Velocity of the Earth: $\frac{2\pi*149.6*10^9meters}{1 year}$

Estimate the centripetal acceleration of the earth in relation to the sun.

Possible Answers:

$6.22*10^{-3}\frac{m}{s^2}=a_{cent}$

$5.97*10^{-3}\frac{m}{s^2}$

$4.44*10^{-3}\frac{m}{s^2}=a_{cent}$

$8.67*10^{-3}\frac{m}{s^2}=a_{cent}$

$4.67*10^{-3}\frac{m}{s^2}$

Correct answer:

$5.97*10^{-3}\frac{m}{s^2}$

Explanation:

$a_{cent}=\frac{v^2}{r}$

The velocity of the earth is $\frac{2\pi*149.6*10^9meters}{1 year}$

Convert to $\frac{meters}{second}$

$\frac{9.40*10^{11}meters}{year}*\frac{1 year}{365.25 days}*\frac{1 day}{24 hours}*\frac{1 hour}{3600 seconds}=\frac{2.99*10^4m}{s}$

Plug in values:

$a_{cent}=\frac{\left(\frac{2.99*10^4m}{s}\right)^2}{149.6*10^9m}$

$5.97*10^{-3}\frac{m}{s^2}=a_{cent}$

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AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1041 : Ap Physics 1

Example Question #51 : Circular And Rotational Motion

Example Question #21 : Centripetal Force And Acceleration

Example Question #22 : Centripetal Force And Acceleration

Example Question #54 : Circular And Rotational Motion

Example Question #23 : Centripetal Force And Acceleration

Example Question #102 : Circular, Rotational, And Harmonic Motion

Example Question #103 : Circular, Rotational, And Harmonic Motion

Example Question #24 : Centripetal Force And Acceleration

Example Question #21 : Centripetal Force And Acceleration

All AP Physics 1 Resources

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