A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?
Solution:
The linear velocity, v and the angular velocity \omega are related by the equation
v=r\omega \ \text{or} \ \omega=\frac{v}{r}
From the given problem, we are given the following values: r=0.420 \ \text{m} and v=32.0 \ \text{m/s}. Substituting these values into the formula, we can directly solve for the angular velocity.
(a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4×106 m at its equator, what is the linear velocity at Earth’s surface?
Solution:
Part A
The period of a rotating body is the time it takes for 1 full revolution. The Earth rotates about its axis, and complete 1 full revolution in 24 hours. Therefore, the period is
The angular velocity \omega is the rate of change of an angle,
\omega = \frac{\Delta \theta}{\Delta t},
where a rotation \Delta \theta takes place in a time \Delta t.
From the given problem, we are given the following: \Delta \theta = 2\pi \text{radian} = 1 \ \text{revolution}, and \Delta t =24\ \text{hours} = 1440 \ \text{minutes}= 86400 \ \text{seconds}. Therefore, the angular velocity is
The linear velocity v, and the angular velocity \omega are related by the formula
v = r \omega
From the given problem, we are given the following values: r=6.4 \times 10^{6} \ \text{meters}, and \omega = 7.27 \times 10^{-5}\ \text{radians/second}. Therefore, the linear velocity at the surface of the earth is
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)
Solution:
We are given the following: v_{0}=1000 \ \text{km/h}, v_{f}=0 \ \text{km/h}, \Delta t = 1.1 \ \text{s}.
The acceleration is computed as the change in velocity divided by the change in time.
(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4 \times 10^{4} N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?
Solution:
Considering the direction of motion as the positive direction, we are given the following: T=2.4 \times 10^4 \ \text{N}, f=-650 \ \text{N}, and mass, m=2100 \ \text{kg}.
Part A. The magnitude of the acceleration can be computed using Newton’s Second Law of Motion.
\begin{align*}
\Sigma F & =ma \\
2.4\times 10^4 \ \text{N}-650 \ \text{N} & = 2100 \ \text{kg}\times a \\
23350 & = 2100 a \\
\frac{23350}{2100} & = \frac{\cancel{2100} a}{\cancel{2100}} \\
a & = \frac{23350}{2100} \\
a & = 11 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part B. The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s2. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.
Solution:
Since the rockets are off, the only force acting on the sled is the friction f. This force is against the direction of motion. By using Newton’s Second Law of Motion, we have.
In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?
Solution:
We can isolate the mower and expose the forces acting on it. This is the free-body diagram.
There are two forces acting on the mower in the horizontal directions:
\textbf{F}:This is the force exerted by the person on the mower, and it is going to the right. This is the first unknown in the problem. We treat this as a positive force since it is directed to the right. The value of this force is F=51 \ \text{N}
\textbf{f}: This is the friction force directed opposite the motion of the mower. We treat this as a negative force because it is directed to the left. The value of this force is f=24 \ \text{N}.
Part A. The third force in the figure is the net force, F_{net}. This is the vector sum of the forces F and f. That is
The person should exert a force of 75 N to produce a net force of 51 N.
Part B. When the force F is removed, the friction is now the only force acting on the mower. The friction is acting opposite the direction of motion. The direction of motion is indicated by the blue arrow in the figure below, and the friction force is the red arrow.
Using Newton’s Second Law, we can solve for the deceleration of the mower.
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s2. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.
Solution:
We are given the following: \sum F = 50.0 \ \text{N}, and a=0.893 \ \text{m/s}^{2}.
Part A. We can solve for the mass, m by using Newton’s second law of motion.
\begin{align*}
\sum F & = ma \\
50.0 \ \text{N} & = m \left( 0.893 \ \text{m/s}^{2} \right) \\
m & = \frac{50.0 \ \text{N}}{0.893 \ \text{m/s}^{2}} \\
m & = 56.0 \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part B. The measured acceleration is equal to the sum of the accelerations of the astronauts and the ship. That is
a_{measured}=a_{astronaut}+a_{ship}
If a force acting on the astronaut came from something other than the spaceship, the spaceship would not undergo a recoil. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.
Solution:
Part A
Consider Figure A.
We are interested in two positions. Position 1 is where the coin is dropped. At this position, the coin is 300 m above the ground, the time is 0 s, and the velocity is 10.0 m/s upward.
Position 2 is the highest point of the coin reaches. At this position, the velocity is equal to 0 m/s.
Position 1 is the initial position and position 2 is the final position. Solve for the value of y2.
\therefore The maximum height reached by the coin is about 305 meters from the ground.
Part B
We do not know the position 4 seconds after the coin has been released, the answer can be above or below the initial point. We can actually use one of the kinematical equations to solve for the final position given the time. Here, the initial position is the point of release and the final position is the point of interest at 4.00 seconds after release.
\therefore The coin is at a height of 262 meters above the ground 4.00 seconds after release. That is, the coin is already dropping and it is already below the release point.
Solving for the velocity 4.00 seconds after release considering the same initial and final position.
\therefore The coin has a velocity of 29.2 m/s directed downward 4.00 seconds after it is released. This confirms that the coin is indeed moving downwards at this point.
Part C
Considering figure C, we have two positions. Position 1 is the point of release 300 m above the ground with a velocity of 10 m/s upward. This is time 0 s.
The second position is at the ground where y=0 m. We are interested at the time in this position.
Considering position 1 as the initial position and position 2 as the final position.
A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10−5 s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
Solution:
Part A
For this part, we shall consider Figure A.
We will be considering the two positions as shown. The first position is when the ball is dropped from a height of 1.50 meters. For this position, we know that y1=1.50 m, t1=0 s, and vy1=0 m/s.
Position 2 is immediately after the ball hits the floor. For this position, we do not the time elapse and the velocity but we know that the height is zero. That is y2=0 m.
Position 1 is the initial position and position 2 is the final position. Solving for vy2, we have
\therefore The steel ball has a velocity of about 5.42 m/s directed downward when it strikes the floor.
Part B
Figure B shows the two positions we are interested in to solve for this part.
Position 1 is at the floor immediately just after the ball hits it. At this initial position, we have y1=0 m, and t1= 0 s. We do not know the velocity at this point.
At position 2, the ball bounced back to its second peak. We know that at the peak of a free falling body, the velocity is zero. So, for this final position, we have y2=1.45 m, and vy2=0 m/s. We do not know the time at this position.
Position 1 is the initial position while position 2 is the final position. Solving for the initial velocity we have
\therefore The steel ball has a velocity of about 5.33 m/s directed upward immediately after it leaves the floor.
Part C
From our answers in Part A and Part B, we have a change in velocity from -5.42 m/s to 5.33 m/s. So, in this case the initial velocity is v1=-5.42 m/s and the final velocity is v2=5.33 m/s. We can compute for the acceleration:
The period of compression happens when the ball has a velocity of -5.42 m/s until it reaches 0 m/s. We shall solve for the change in displacement for this two given velocities. The initial velocity is -5.42 m/s and the final velocity is 0 m/s. The acceleration during this period is the one solved in Part C, a=1.34×105 m/s2.
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