PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 6, Problem 92P

(a)

To determine

The speed at which the car reach the top of the loop.

(a)

Expert Solution
Check Mark

Answer to Problem 92P

The speed at which the car reach the top of the loop is 19.8m/s.

Explanation of Solution

The mass of the roller coaster car is 988kg. The diameter of the circular loop is 20.0m. The car is at rest initially at a height of 40.0m.

Write the formula for the conservation of energy for the car.

Ki+Ui=Kf+Uf (I)

Here, Ki is the initial kinetic energy, Ui is the initial potential energy, Kf is the final kinetic energy, Uf is the final potential energy

Consider the initial state to be at the top at rest and final stage to be at the top of the loop. The initial kinetic energy is thus zero.

Re-write equation (I).

0+mghi=12mvf2+mghf

Here, m is the mass of the car, g is the acceleration due to gravity, hi is the initial height of the car, vf is the final velocity of the car, hf is the final height of the car.

Re-write the above equation to get an expression for vf.

vf=2mg(hihf)m=2g(hihf) (II)

Conclusion:

Substitute 988kg for m, 40.0m for hi, 20.0m for hf, 9.80m/s2 for g in equation (III).

vf=2(9.80m/s2)(40.0m20.0m)=19.8m/s

The speed at which the car reach the top of the loop is 19.8m/s.

(b)

To determine

The force exerted by the track on the car at the top of the loop.

(b)

Expert Solution
Check Mark

Answer to Problem 92P

The force exerted by the track on the car at the top of the loop is 29kN.

Explanation of Solution

The mass of the roller coaster car is 988kg. The diameter of the circular loop is 20.0m. The car is at rest initially at a height of 40.0m.

At the top of the loop, the normal force and the weight of the car points downward and provide the centripetal force.

Write the equation for forces at the top of the loop in vertical direction.

mv2r=N+mg

Here, m is the mass of the car, g is the acceleration due to gravity, v is the velocity of the car at the top of the loop, N is the normal force by the track on the car, r is the radius of the loop.

Re-write the above equation to get an expression for N.

N=m(v2rg)

Conclusion:

Substitute 988kg for m, 19.8m/s for v, 20.0m2 for r, 9.80m/s2 for g in equation (III).

N=(988kg)((19.8m/s)220.0m/29.80m/s2)=29051N29kN

The force exerted by the track on the car at the top of the loop is 29kN.

(c)

To determine

The minimum height for the release of the car so that the car does not lose contact with the track.

(c)

Expert Solution
Check Mark

Answer to Problem 92P

The force exerted by the track on the car at the top of the loop is 29.0kN.

Explanation of Solution

The mass of the roller coaster car is 988kg. The diameter of the circular loop is 20.0m. The car is at rest initially at a height of 40.0m.

At the top of the loop, the normal force and the weight of the car points downward and provide the centripetal force.

Write the equation for forces at the top of the loop in vertical direction.

mv2r=N+mg

Here, m is the mass of the car, g is the acceleration due to gravity, v is the velocity of the car at the top of the loop, N is the normal force by the track on the car, r is the radius of the loop.

At the minimum height for the release of the car so that the car does not lose contact with the track, the normal force is zero.

Re-write the above equation.

mv2r=N+mg (I)

From section (a) write the formula for the velocity at the top of the loop.

v=2g(hihf) (II)

Here, hi is the initial height of the car, hf is the final height of the car.

Substitute equation (II) in equation (I).

m(2g(hihf))r=mg

Re-write the above equation to get an expression for hi.

hi=r+2hf2

Conclusion:

Substitute, 20m for hf, 20.0m2 for r in equation (III).

hi=20m/2+2(20m)2=10m+40m2=25m

The minimum height for the release of the car so that the car does not lose contact with the track is 25m.

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Chapter 6 Solutions

PHYSICS

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