e) A cart slides down a frictionless inclined track to a circular loop of radius

Question

e) A cart slides down a frictionless inclined track to a circular loop of radius R- 13 m. In order for the cart to negotiate the loop safely, the normal force acting on the cart at the top of the loop, due to the track, must be at least equal to the cartl's weight. (Note: This is different from the conditions needed to just negotiate the loop.) 1) What must be the minimum speed IVminl of the cart at the top of the loop? m/s /s Submit Help 2) How high h above the top of the loop must the cart be released? m Submit Help 3) When the car is descending vertically in the loop (point (c) in the picture), what is its speed lvl? m/s Submit Help

Explanation / Answer

1.)

here by using the formula

m * v^2 / r = m * g

v = sqrt( r * g)

v = sqrt(13 * 9.8 )

v = 11.28 m/s

2.)

PE(inital) = KE(final) + PE(final)

m * g * h = 0.5 * m * v^2 + m * g * H

here H = 2 * R = 2 * 13 = 26 m

9.8 * h = 0.5 * 11.28^2 + 9.8 * 26

h = 32.49 m

c)

0.5 * m * v^2 = m * g * h

v = sqrt( 2 * g * h)

v = sqrt( 2 * 9.8 * 13)

v = 15.96 m/s

d)

F * d = PE(initial)

m * a * d = m * g * h

a = g * h / d

a = 9.8 * 32.49 / 10

a = 31.84 m/s^2

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