**PHY 1351**

Hour Exam 2

October 17, 2001

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** Conceptual Questions:
1. Q6.4 Why does mud fly off a rapidly turning automobile tire? **

For mud to move in a circle, there must be a (net) force on it -- directed toward the center of the circle. The value of that force is F

_{c}= m v^{2}/ r. As the velocity (or speed) increases, that force must increase if the piece of mud is to continue to move in a circle. As the velocity increases, the force that holds the mud to the wheel reaches its limit and the mud can no longer go around in a circle. When that limit is reached, the mud separates from the tire.

2. Q6.8 Describe a situation in which a car driver can have a centripetal acceleration
but no tangential acceleration.

Uniform Circular Motion (UCM) is just such a situation. There will always be a centripetal acceleration in circular motion because of the changing direction. If the motion is uniform -- if the speed is constant -- there will be no tangential acceleration.

3. Q7.4 Can the kinetic energy of an object be negative?

No. Kinetic energy is given by KE = (

^{1}/_{2}) m v^{2}. The mass m is always positive and v^{2}is positive so KE must always be positive (of course, if v = 0 then KE = 0). KE cannotbe negative.

4. Q7.14 An older model car accelerates from 0 to a speed v in 10 s. A newer,
more powerful sports car accelerates from 0 to 2 v in the same time period.
What is the ratio of powers expended by the two cars?

Assume the two cars have the same mass.

The car going 2 v has

fourtimes the Kinetic Energy. If it acquired this much energy in the same time, then the power supplied to it is alsofourtimes as much.

5. Q8.10 What would the curve of U versus x look like if a particle were in
a region of neutral equilibrium?

Neutral equilibrium means zero net force. The force is the slope of the U vs x curve. That means the slope is zero so the curve is a straight, horizontal line.

**1. (6.63) An amusement park ride consists of a large vertical cylinder that
spins about its axis fast enough that any person inside is held up against the
wall when the floor drops away. The coefficient of static friction between the
person and the wall is µ _{s}, and the radius of the cylinder is
R.
Show that the maximum period of revolution necessary to keep the person from
falling is T = (4 ^{2}
R µ_{s} /g)^{1/2} or T = SQRT(4 ^{}^{2}
R µ_{s} /g) . **

** **

** **F_{x} = - F_{n}
= - m v^{2} / R = - F_{c}

F_{n} = m v^{2} / R

** **F_{y} = F_{f} -
m g = 0 = m a_{y}

F_{f} = m g

F_{f} = F_{n}

F_{f} = F_{n} = ( m v^{2} / R)
= m g

( m v^{2} / R)
= m g

v = C / T = 2 R / T

v^{2} = 4 ^{2} R^{2}
/ T^{2}

( [4 ^{2} R^{2}
/ T^{2}] / R) = g

T^{2} = 4 ^{2} R / g

T = [ 4 ^{2} R / g ]^{1/2}

**2. (6.21) A roller-coaster vehicle is on the track shown below.
(a) What is the maximum speed the vehicle can have at B and still remain on
the track?
(b) What must be the vehicle’s speed at point A to reach point B with the
speed you just calculated?**

To "still remain on the track" means the normal force F_{n} has just
gone to zero, F_{n} = 0.

With the normal force equal to zero, there is only the weight mg available
to supply the centripetal force F_{c},

F_{c} = F_{net}

m v^{2} / r = m g

v^{2} / r = g

v^{2} = g r = (9.8 m / s^{2} ) (15 m) = 147
m^{2}/s^{2}

**v = 12.12 m /s**

**(b) What must be the vehicle’s speed at point A to reach
point B with the speed you just calculated?**

E_{A} = E_{B}

K_{A} + U_{A} = K_{B} + U_{B}

( ^{1}/_{2} ) M v_{A}^{2} +
0 = ( ^{1}/_{2} ) M (12.12 ^{m}/_{s})^{2}
+ M (9.8 m/s^{2}) (20.0 m)

( ^{1}/_{2} ) v_{A}^{2} = (
^{1}/_{2} ) (12.12 ^{m}/_{s})^{2} +
(9.8 m/s^{2}) (20.0 m)

( ^{1}/_{2} ) v_{A}^{2} = (
^{1}/_{2} ) (147 m^{2}/s^{2} ) + ( 196 m^{2}/s^{2}
)

( ^{1}/_{2} ) v_{A}^{2} = (
73.5 m^{2}/s^{2} ) + ( 196 m^{2}/s^{2}
)

( ^{1}/_{2} ) v_{A}^{2} = (
269.5 m^{2}/s^{2} )

v_{A}^{2} = 539 m^{2}/s^{2}

**v _{A} = 23.2 ^{m}/_{s}**

** 3. (4 ed; 7.4) A block of mass 0.60 kg slides 6.0 m down a frictionless
ramp inclined at 20o to the horizontal. It then travels on a rough horizontal
surface where k = 0.50.
**

**(a) What is the speed of the block at the end of the incline?
(b) What is its speed after traveling 1.00 m on the rough surface?
(c) What distance does it travel on this horizontal surface before stopping?
**

Onthe frictionlessincline, the normal force doesno work, W_{n}= 0. Only the force of gravity does any work,W

_{g}= F s cos = ( m g ) (6.0 m) cos 70^{o}= (0.60 kg) (9.8 m/s^{2}) (6.0 m) (0.342) = W_{g}= 12.1 JNotice that this has the cosine of 70

^{o}rather than of 20^{o}. In writing W = F s cos , the angle is the anglebetweenthe force and the displacement.W

_{net}= W_{g}+ W_{n}= 12.1 JW

_{net}= KE = KE - KE_{i}= KEW

_{net}= (^{1}/_{2}) m v^{2}(

^{1}/_{2}) m v^{2}= (^{1}/_{2}) (0.60 kg) v^{2}= 12.1 Jv

^{2}= 40.22 m^{2}/ s^{2}

v = 6.34 m / sThis is the speed at the bottom of the incline and the beginning of the rough horizontal plane that has friction.

From the force diagram, you can see that

F

_{n}= m g = (0.60 kg) (9.8 m/s^{2}) = 5.88 NF

_{f}= F_{n}= (0.5) (5.88 N) = 2.94 N, pointing to theleftAlong this rough, horizontal plane,

onlythis friction force F_{f}does any work. By the time the block has moved a distance of 1.0 m along the plane, the net work done on the block isW

_{net}= W_{f}= - (2.94 N) (1 m) = - 2.94 JW

_{net}= KE = KE_{1.0 m}- KE_{o}= KE_{1.0 m}- 12.1 J = - 2.9 JKE

_{1.0 m}= 12.1 J - 2.9 J = 9.2 JKE

_{1.0 m}= 9.2 J = (^{1}/_{2}) m v^{2}= (^{1}/_{2}) (0.60 kg) v^{2}v

^{2}= 30.7 m^{2}/ s^{2}

v = 5.5 m / sNow, how far does it travel before it completely stops? At that point, KE

_{f}= 0. The work done by friction in moving this distance X isW

_{net}= W_{f}= - (2.94 N) ( X_{f})And this is just equal to the change in KE, from its initial value of 12.1 J at the bottom of the inclined plane until it goes to zero,

- (2.94 N) X

_{f}= - 12.1 J

X_{f}= 4.11 m

We will measure the gravitational potential energy U_{g}
from the "final" position of the 5.0-kg mass.

E_{i} = 0 + m_{2} g h_{2}

E_{i} = 0 + (5.0 kg) (9.8 m/s^{2}) (2.0 m)

E_{i} = 98.0 J

This amount of energy goes into the **kinetic energy** of the system --
the kinetic energy of **both** masses --** and** the heat associated with
the work due to friction.

K_{f} = (^{1}/_{2}) (8 kg) v^{2} = (4 kg)
v^{2}

U_{f} = 0

E_{f} = E_{i} + W_{f}

W_{f} = - F_{f}s

F_{f} =
F_{n}

F_{n} = m_{1} g

F_{n} = (3.0 kg) (9.8 m/s^{2})

F_{n} =29.4 N

F_{f} = (0.30) (29.4 N) = 8.82 N

W_{f} = - (8.82 N) (2.0 m)

W_{f} = - 17.6 J

E_{f} = 98.0 J - 17.6 J = 80.4 J

E_{f} = K_{f} = (1/2)(m_{1} + m_{2}) v^{2}
= (1/2) (3 kg + 5 kg) v^{2}

E_{f} = K_{f} =(4 kg) v^{2} = 80.4 J

v^{2}= 20.1 m^{2}/s^{2}

**v = 4.5 m/s**

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