Motion - AP Physics C: Electricity and Magnetism

Card 0 of 880

Question

A wind catcher is created by attaching four plastic bowls of mass each to the ends of four lightweight rods, which are then secured to a central rod that is free to rotate in the wind. The four lightweight rods are of lengths , , , and .

Calculate the moment of inertia of the four bowls about the central rod. You may assume to bowls to be point masses.

Answer

The moment of inertia for a point mass is .

To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that

$I_{total}=I_{1}+I_{2}+I_{3}+I_{4}$

$I_{total}=m_{1}R_{1}^{2}+m_{2}R_{2}^{2}+m_{3}R_{3}^{2}+m_{4}R_{4}^{2}$

The masses of the bowls are all equal in this problem, so this simplifies to

$I_{total}=m (R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2})$

Plugging in and solving with numerical values,

$I_{total}=(0.6: kg) [(0.8: m)^{2}+(0.6: m)^{2}+(0.4: m)^{2}+(0.2: m)^{2}]=0.72: kg\cdot m^{2}}$

Compare your answer with the correct one above

Question

A long uniform thin rod of length has a mass of .

Calculate the moment of rotational inertia about an axis perpendicular to its length passing through a point from one of its ends.

Answer

For a long thin rod about its center of mass,

$I_{cm}=\frac{1}{12}mL^{2}$

According to the parallel axis theorem,

$I_{//}=I_{cm}+md^{2}$

where is defined to be the distance between the center of mass of the object and the location of the axis parallel to one through the center of mass. For this problem, is the difference between the given distance and half the length of the rod.

Combining the above,

$I=\frac{1}{12}mL^{2}+m(\frac{1}{2}L-l)^{2}$

Inserting numerical values,

$I=\frac{1}{12}(12)(1.82)^{2}+(12)(\frac{1.82}{2}-.036)^{2}=6.9: kg\cdot m^{2}$

Compare your answer with the correct one above

Question

The moment of inertia of a long thin rod about its end is determined to be .

What is the new value if the mass and length of the rod are both reduced by a factor of ?

Answer

The moment of inertia for a long uniform thin rod about its end is given by

$I_{rod,end}=\frac{1}{3}mL^{2}$

Reducing the mass and the length by a factor of four introduces the following factors into the equation,

$I_{new}=\frac{1}{3}(\frac{1}{4}m)(\frac{1}{4}L)^{2}=\frac{1}{3}(\frac{1}{4})m(\frac{1}{64})L^{2}$

Simplifying,

$I_{new}=(\frac{1}{64})\frac{1}{3}mL^{2}=\frac{1}{64}I_{0}$

Compare your answer with the correct one above

Question

An art sculpture comprised of a long hollow painted cylinder is designed to pivot about an axis located along its long edge. The cylinder has mass , radius , and length .

Calculate the moment of inertia of the sculpture about this axis.

Answer

The moment of inertia for a hollow cylinder about its center is given by

$I_{cyl,hollow}=mR^{2}$

The parallel axis is applied here because the pivot axis is located on the outside edge of the cylinder, a distance from the center of mass:

$I=I_{cm}+md^{2}=mR^{2}+mR^{2}=2\cdot mR^{2} = 2\cdot234: kg\cdot(2.2: m)^{2}=2,265: kg\cdot m^{2}$

Incidentally, for this problem, the length does not contribute to the moment of inertia.

Compare your answer with the correct one above

Question

A meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. A force of is applied perpendicularly to its length at a distance from the nailed end.

Calculate the resulting angular acceleration experienced by the meter stick.

Answer

For a net torque applied to an object free to rotate,

$\tau _{net}=I\alpha$

The net torque for this problem is supplied by the force applied at a distance from the pivot. The torque of a force is calculated by

$\tau =Fr\sin, \theta$

The moment of inertia for a long uniform thin rod about its end is

$I_{rod,end}=\frac{1}{3}mL^{2}$

Combining above,

$Fr\sin90^{\circ}=(\frac{1}{3}mL^{2})\alpha$

Solving for the angular acceleration,

$\alpha =\frac{Fr\sin, 90^{\circ}}{\frac{1}{3}mL^{2}}=\frac{16N\cdot0.67: m\cdot1}{\frac{1}{3}\cdot2, kg\cdot(1:m)^{2}}=16: rad/s^{2}$

Compare your answer with the correct one above

Question

A meter stick is secured at one end so that it is free to rotate in a vertical circle. It is held perfectly horizontally, and released.

Calculate the instantaneous angular acceleration the moment it is released.

Answer

For a net torque applied to an object free to rotate,

$\tau _{net}=I\alpha$

The net torque for this problem is supplied by the force applied at a distance from the pivot. The torque of a force is calculated by

$\tau =Fr\sin, \theta$

The moment of inertia for a long uniform thin rod about its end is

$I_{rod,end}=\frac{1}{3}mL^{2}$

Combining above,

$Fr\sin90^{\circ}=(\frac{1}{3}mL^{2})\alpha$

where the force causing the torque is the meter stick's weight, which is applied at the center of mass of the meter stick - a distance of half the length away from its end.

Solving for the angular acceleration,

$\alpha =\frac{Fr\sin, 90^{\circ}}{\frac{1}{3}mL^{2}}=\frac{W\cdot\frac{1}{2}L\cdot \sin, 90^{\circ}}{\frac{1}{3}mL^{2}}$

Solving numerically,

$\alpha =\frac{(2: kg\cdot 9.8: m/s^{2})\cdot 0.5: m\cdot 1}{\frac{1}{3}\cdot 2, kg\cdot (1:m)^{2}}=15: rad/s^{2}$

Compare your answer with the correct one above

Question

The velocity (in meters per second) of a moving particle is given by the following function:

If the particle's initial position is 0m, what is the position of the particle after two seconds?

Answer

To solve this problem you need to obtain a function of position with respect to time. For that, you need to understand velocity as the rate of change of displacement with respect to time. In other words, velocity is "how fast" (i.e. how much time it takes) an object changes position (remember displacement is the change in position). This means that velocity is the derivative of displacement with respect to time.

Therefore, to obtain a function of position with respect to time you need to take the antiderivative of the velocity function, so we integrate:

$x(t) = \int(3t^2)dt = \frac{3t^3}{3} + x_{0}= t^3 + x_{0}$

Here, $x_{0}$ is a constant that represents the initial position of the particle. We know that the initial position of the particle is 0m, so our function is:

Therefore, after two seconds have passed, we have t = 2s and

Note: we know that position is given in meters since the question specified that velocity is measured in meters per second.

Compare your answer with the correct one above

Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

Compare your answer with the correct one above

Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

Compare your answer with the correct one above

Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the the approximate mass of the guillotine blade?

Answer

The force of gravity on the blade is , which is the same as $400\frac{kg\cdot m}{s^2}$

This unit relationship comes from Newton's second law.

is the mathematical expression of Newton's second law. The units for force must be a product of the units for mass and the units for acceleration.

Solve the expression by plugging in known values.

$400 \frac{kg\cdot m}{s^2}=m \cdot 9.8\frac{m}{s^2}$

$\frac{400}{9.8}kg=mass\approx41kg$

Compare your answer with the correct one above

Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the tension on the rope pulling the blade, while it is accelerating into position?

Answer

The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.

$F_{T}-F_{g}=ma$

Rearrange the equation.

$F_{T}=ma+F_{g}$

Plug in known values.

$F_T=41kg\cdot2\frac {m}{s^2} + 400N$

Compare your answer with the correct one above

Question

An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation $\small y=ax^2+bx +c$ . Its motion in the horizontal direction is described by the equation $\small x=v_{x}t$ . What is the object's velocity is the direction in terms of its horizontal position $\small x$ ?

Answer

The y velocity is thetime derivative of the $\small y$ position, and not the $\small x$ derivative. In order to find it, use the chain rule:

$\frac{dy}{dt}=\left(\frac{dy}{dx}\right) \frac{dx}{dt}$

$\frac{dy}{dx}= 2ax + b$

Of course, $\frac{dx}{dt} =v_{x}$

Compare your answer with the correct one above

Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, what is true about the two objects at time $\small t=5s$ ?

Answer

Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.

Compare your answer with the correct one above

Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, which object has traveled a greater distance from its starting position when $\small t=5s$ ?

Answer

Since this is a graph of velocity vs. time, its integral is distance travelled. We can estimate the integral by looking at the area under the curves. Since Bbject 1 has a greater area under its velocity curve, it has covered a greater distance. Its velocity is greater that Object 2's for the entire time, so it makes sense that it will travel farther.

Compare your answer with the correct one above

Question

Atwood's machine consists of two blocks connected by a string connected over a
pulley as shown. What is the acceleration of the blocks if their masses are and .

Assume the pulley has negligible mass and friction.

Answer

From the force diagram above, we can see that tension is pulling up on both sides of the string and gravity is pulling down on both blocks. With this information we can write 2 force equations:

If we add the two equations together, we get:

where

Solving for , we get

Compare your answer with the correct one above

Question

A car undergoes acceleration according to the given function. What distance has the car traveled after three seconds?

Answer

We can determine the velocity by taking the second integral of acceleration for the time interval of 0s to 3s.

$\int a(t)dt = v(t)$

$\int_{0}^{3}\int a(t)dt^2 = x(3) - x(0)$

Solve for the first integral.

$\int_{0}^{3}\int a(t)dt^2 =\int_{0}^{3}(6t^2)\int dt^2=\int_{0}^{3}(2t^3)dt$

Solve for the second integral, using the time interval.

$\int_{0}^{3}(2t^3)dt = \frac{(3)^4}{2}-\frac{(0)^4}{2} = \frac{81}{2} = 40.5m$

Compare your answer with the correct one above

Question

A car undergoes acceleration according to the given function. How fast is the car moving after four seconds?

Answer

We can find the car's velocity by taking the integral of the acceleration function during the given time interval.

$v(t) = \int a(t),dt$

$v(4) = \int_{0}^{4} a(t),dt = \int_{0}^{4} (6t^2),dt$

Solve the integral for the time interval of 0s to 4s. This will give us the final velocity.

$\int_{0}^{4} (6t^2),dt = 2(4)^3+2(0)^3 = 128\ \frac{m}{s}$

Compare your answer with the correct one above

Tap the card to reveal the answer