L>1351Key2F01PHY 1351 Hour Exam 2 October 17, 2001 Return to PHY 1350"s Home Page

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

For mud to relocate in a circle, there need to be a (net) force on it -- directed toward the facility of the circle. The worth of that pressure is Fc = m v2/ r. As the velocity (or speed) rises, that force should increase if the item of mud is to proceed to relocate in a circle. As the velocity boosts, the pressure that holds the mud to the wheel reaches its limit and also the mud have the right to no longer go approximately in a circle. When that limit is reached, the mud separates from the tire.

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

Unidevelop Circular Motion (UCM) is simply such a situation. Tbelow will certainly always be a centripetal acceleration in circular motion bereason of the transforming direction. If the activity is unidevelop -- if the speed is continuous -- there will be no tangential acceleration.

3. Q7.4 Can the kinetic energy of a things be negative?

No. Kinetic power is given by KE = (1/2) m v2. The mass m is constantly positive and v2 is positive so KE should constantly be positive (of course, if v = 0 then KE = 0). KE have the right to not be negative.

4. Q7.14 An older version auto accelerates from 0 to a speed v in 10 s. A newer, even more effective sporting activities vehicle increases 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 actually the same mass.

The car going 2 v has actually four times the Kinetic Energy. If it acquired this a lot energy in the same time, then the power provided to it is additionally four times as a lot.

5. Q8.10 What would the curve of U versus x look prefer if a ppost were in an area of neutral equilibrium?

Neutral equilibrium suggests zero net pressure. The pressure is the slope of the U vs x curve. That indicates the slope is zero so the curve is a right, horizontal line.

Calculational Questions:

1. (6.63) An amusement park ride consists of a huge vertical cylinder that spins about its axis fast enough that any type of person inside is held up against the wall when the floor drops away. The coreliable of static friction between the perkid and also the wall is µs, and the radius of the cylinder is R. Show that the maximum duration of radvancement important to keep the perchild from falling is T = (4 2 R µs /g)1/2 or T = SQRT(4 2 R µs /g) .

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Fx = - Fn = - m v2 / R = - Fc

Fn = m v2 / R

Fy = Ff - m g = 0 = m ay

Ff = m g

Ff = Fn

Ff = Fn = ( m v2 / R) = m g

( m v2 / R) = m g

v = C / T = 2 R / T

v2 = 4 2 R2 / T2

( <4 2 R2 / T2> / R) = g

T2 = 4 2 R / g

T = < 4 2 R / g >1/2


2. (6.21) A roller-coaster car is on the track displayed below. (a) What is the maximum speed the automobile can have actually at B and also still remain on the track? (b) What need to be the vehicle’s rate at point A to reach point B with the speed you simply calculated? To "still reprimary on the track" implies the normal force Fn has simply gone to zero, Fn = 0.

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With the normal force equal to zero, tright here is just the weight mg available to supply the centripetal pressure Fc,

Fc = Fnet

m v2 / r = m g

v2 / r = g

v2 = g r = (9.8 m / s2 ) (15 m) = 147 m2/s2

v = 12.12 m /s

(b) What must be the vehicle’s speed at point A to reach allude B via the rate you simply calculated?EA = EBKA + UA = KB + UB

( 1/2 ) M vA2 + 0 = ( 1/2 ) M (12.12 m/s)2 + M (9.8 m/s2) (20.0 m) ( 1/2 ) vA2 = ( 1/2 ) (12.12 m/s)2 + (9.8 m/s2) (20.0 m) ( 1/2 ) vA2 = ( 1/2 ) (147 m2/s2 ) + ( 196 m2/s2 ) ( 1/2 ) vA2 = ( 73.5 m2/s2 ) + ( 196 m2/s2 ) ( 1/2 ) vA2 = ( 269.5 m2/s2 ) vA2 = 539 m2/s2 vA = 23.2 m/s 3. (4 ed; 7.4) A block of mass 0.60 kg slides 6.0 m dvery own a frictionless ramp inclined at 20o to the horizontal. It then travels on a stormy horizontal surconfront wbelow k = 0.50.

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

On the frictionless incline, the normal force does no work, Wn= 0. Only the force of gravity does any kind of work,

Wg = F s cos = ( m g ) (6.0 m) cos 70o = (0.60 kg) (9.8 m/s2) (6.0 m) (0.342) = Wg = 12.1 J

Notice that this has the cosine of 70o rather than of 20o. In writing W = F s cos , the angle is the angle between the force and also the displacement.

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Wnet = Wg + Wn = 12.1 J

Wnet = KE = KE - KEi = KE

Wnet = (1/2) m v2

(1/2) m v2 = (1/2) (0.60 kg) v2 = 12.1 J

v2 = 40.22 m2 / s2

v = 6.34 m / s

This is the rate at the bottom of the incline and the beginning of the unstable horizontal aircraft that has actually friction.

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From the pressure diagram, you have the right to check out that

Fn = m g = (0.60 kg) (9.8 m/s2) = 5.88 N

Ff = Fn = (0.5) (5.88 N) = 2.94 N, pointing to the left

Alengthy this rough, horizontal aircraft, only this friction pressure Ff does any kind of work-related. By the time the block has actually relocated a distance of 1.0 m along the plane, the net occupational done on the block is

Wnet= Wf = - (2.94 N) (1 m) = - 2.94 J

Wnet = KE = KE1.0 m - KEo = KE1.0 m - 12.1 J = - 2.9 J

KE1.0 m = 12.1 J - 2.9 J = 9.2 J

KE1.0 m = 9.2 J = (1/2) m v2 = (1/2) (0.60 kg) v2

v2 = 30.7 m2 / s2

v = 5.5 m / s

Now, exactly how much does it take a trip prior to it totally stops? At that suggest, KEf = 0. The work done by friction in relocating this distance X is

Wnet= Wf = - (2.94 N) ( Xf )

And this is simply equal to the change in KE, from its initial value of 12.1 J at the bottom of the inclined plane till it goes to zero,

- (2.94 N) Xf = - 12.1 J

Xf = 4.11 m

4. (8.31) The coefficient of friction between the 3.0-kg mass and surconfront shown below is 0.30. The device starts from rest. What is the rate of the 5.0-kg mass once it has actually fallen 2.0 m?

We will measure the gravitational potential power Ug from the "final" position of the 5.0-kg mass.

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Ei = Ki + Ui Ei = 0 + m2 g h2

Ei = 0 + (5.0 kg) (9.8 m/s2) (2.0 m)

Ei = 98.0 J

This amount of energy goes right into the kinetic energy of the system -- the kinetic energy of both masses -- and the warmth associated through the occupational as a result of friction.

Kf = (1/2) m1 v2 + (1/2) m2 v2 = (1/2) (m1 + m2) v2

Kf = (1/2) (8 kg) v2 = (4 kg) v2