Kinematics (a lot, but most worked out & conceptual)

[clairez93]

Homework Statement

1. A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the acceleration of a passenger at his lowest point during the ride?
a) 5.7 m/s^2 downward
b) 4.1 m/s^2 upward
c) 14 m/s^2 downward
d) 4.1 m/s^2 downward
e) 19 m/s^2 downward

Answer: A

2. A car travels in a due northerly direction at a speed of 55 km/h. The traces of rain on the die windows of the car make an angle of 60 degrees with respect to the horizontal. If the rain is falling vertically with respect to the earth, what is the speed of the rain with respect to the earth?
a) 48 km/h
b) 95 km/h
c) 58 km/h
d) 32 km/h
e) 80 km/h

Answer: B

3. Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A little later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:
a) it has the same position and the same velocity as A
b) it has the same position and the same acceleration as A
c) it has the same velocity and the same acceleration as A
d) it has the same displacement and the same velocity as A
e) it has the same displacement and the same acceleration as A

Answer: E

4. A toy rocket, launched from ground, rises vertically with an acceleration of 20 m/s^2 for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?
a) 1.1 km
b) 0.73 km
c) 1.9 km
d) 0.39 km
e) 1.5 km

Answer: A

5. A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?
a) 17 m
b) 21 m
c) 18 m
d) 27 m
e) 31 m

Answer: A

Homework Equations

The Attempt at a Solution

1. $$\frac{5 rev}{1 min} = \frac{5(2\pi*15)}{60} = 7.85 m/s$$
$$a = \frac{v^{2}}{r} = \frac{7.85^{2}}{15} = 4.11 m/s^{2}$$

Answer says 5.7 m/s^2 though.

2. Not sure how to approach this one at all.

3. Why do the two balls have the same acceleration? If one can pass the other, at a later time, then shouldn't it accelerate faster?

4. $$V_{f} = V_{i} + at$$
$$0 = V_{i} + (20)(6)$$
$$V_{i} = -120$$

$$y = \frac{1}{2}at^{2} + V_{i}t$$
$$= \frac{1}{2}(20)(6)^{2} - 120(6) = -360$$

No answer choices available...

5. $$y_{b} = \frac{1}{2}(-9.8)t^{2} + 20t$$
$$y_{b}(2) = \frac{1}{2}(-9.8)(2)^{2} + 20(2) = 20.4$$
$$y_{b} = \frac{1}{2}(-9.8)t^{2} + 20t + 20.4$$
$$y_{s} = \frac{1}{2}(-9.8)t^{2} + 24t$$
$$\frac{1}{2}(-9.8)t^{2} + 20t + 20.4 = \frac{1}{2}(-9.8)t^{2} + 24t$$
$$4t = 20.4$$
$$t = 5.1$$
$$y = \frac{1}{2}(-9.8)(5.1)^{2} + 24(5.1) = -5.049 m$$

Again, no answer choices there for that...

Any help appreciated, thanks! :]

 

[bakin]

I'm not so sure on the first question. How are you able to figure out the acceleration with just the radius of the circle and revolutions per minute?

 

[Redbelly98]

1. I agree with your 4.1 m/s^2 answer, it seems the "book answer" is wrong. Plus, the acceleration is clearly upward at the lowest point, agreeing with only one answer choice.

3. The reason why ball B passes A has to do with the different initial velocities, not the acceleration.

Hope that helps.

 

[clairez93]

Ah I see, so the acceleration would remain the same for both since only gravity is accelerating them on this hill? Thanks.

Did you happen to catch anything wrong with the other questions?

 

[Redbelly98]

You're welcome. Here's some hints for the other problems.


2. This is a relative velocity problem, and so involves vector addition or subtraction. Review how to find velocity measured in one reference frame (the moving train) that is moving with respect to another (the Earth's surface).


4. The acceleration is not constant for the entire trajectory of the rocket. Instead, it is constant for part of the trajectory. Then it changes, and is a different constant for the remainder of the trajectory.

You'll need to treat the two portions of the trajectory separately, then combine the results.


5. You'll need to account for the fact that the stone is thrown up 2 seconds after the ball. Hint: at t=2s, the stone has y_s=0. You're equation for y_s should agree with that fact, but it doesn't.

 

[clairez93]

For number 4, I recalculated it to get another incorrect answer of 0.734 kilometers.

$$y = \frac{1}{2}at^{2} + V_{i}t = \frac{1}{2}(20)(6^{2}) = 360$$
$$V_{f} = V_{i} + at = (20)(6) = 120$$

For after the second part.

$$V_{f}^2 = V_{i}^2 + 2ad$$
$$0 = 120^{2} + 2(-9.8)d$$
$$d = 734.694 m$$

Help?

 

[jambaugh]

I would also suggest as a matter of work habit that you include all units and make sure they match in an equation. Where you wrote:

$$\frac{5 rev}{1 min} = \frac{5(2\pi*15)}{60} = 7.85 m/s$$
You are equating a rotational speed with a linear speed which are not equivalent quantities (different types of units).

You are implicitly invoking:
$$v = r\omega$$

So you should write
$$v = r\omega = r\cdot \frac{5 rev}{1 min} = (15 m) \cdot \frac{5 rev \cdot 2\pi rad/rev}{60sec} = \frac{15\cdot 5 \cdot 2\pi}{60}\cdot \frac{m}{s}$$
(Note radians are properly dimensionless and can be dropped when necessary.)
When all the units are right then do the arithmetic.

$$v = \frac{5\pi}{2}\frac{m}{s}$$

I realize you may find this a.) a bit tedious and b.) too much typesetting to actually post here but it is good practice on paper and will be less tedious than reworking problems because you miss errors. Many errors become glaring when you do carefully include your units. The units actually help by keeping track of details while you focus on the main equations and concepts (such as $V = r\omega$.) Indeed you can usually infer most of these little relations by paying attention to what the units do.

 

[rl.bhat]

For number 4, I recalculated it to get another incorrect answer of 0.734 kilometers.

$$y = \frac{1}{2}at^{2} + V_{i}t = \frac{1}{2}(20)(6^{2}) = 360$$
$$V_{f} = V_{i} + at = (20)(6) = 120$$

For after the second part.

$$V_{f}^2 = V_{i}^2 + 2ad$$
$$0 = 120^{2} + 2(-9.8)d$$
$$d = 734.694 m$$

Help?

Total height = y + d

 

[rl.bhat]

1. I agree with your 4.1 m/s^2 answer, it seems the "book answer" is wrong. Plus, the acceleration is clearly upward at the lowest point, agreeing with only one answer choice.

3. The reason why ball B passes A has to do with the different initial velocities, not the acceleration.

Hope that helps.

Wheel is a rotating frame of reference. So the passenger sitting in the wheel must experience a centrifugal force. And at the bottom, g is also acting in the downward direction.

 

[rl.bhat]

Ah I see, so the acceleration would remain the same for both since only gravity is accelerating them on this hill? Thanks.

Did you happen to catch anything wrong with the other questions?

In problem 5, when the two balls meet, their displacements must be the same. Before they meet, the first ball is in the air fro t + 2 second and the second ball is in the air for t seconds.

 

[Redbelly98]

1. I agree with your 4.1 m/s^2 answer, it seems the "book answer" is wrong. Plus, the acceleration is clearly upward at the lowest point, agreeing with only one answer choice.

Wheel is a rotating frame of reference. So the passenger sitting in the wheel must experience a centrifugal force. And at the bottom, g is also acting in the downward direction.

The question simply asks for "the acceleration". Since it doesn't mention anything about which frame of reference, shouldn't we assume it's the acceleration as measured in an inertial frame?

In the rotating frame the acceleration would be zero, which isn't one of the answer choices.

However, by subtracting 9.8 m/s^2 we do get the indicated answer of 5.7 m/s^2 downwards. So it seems I am missing something here.