Question

-A roller coaster cart of mass (m1=127 kg) initially sits at rest on a track. Once the motor is activated, the cart is accelerated forward (across level ground) until it reaches the bottom of a giant circular loop with a radius of (r1=17.5 m). At this point, the motor is turned off and the cart must "coast" for the remainder of the ride. Assume the track is frictionless and ignore air resistance. At the top of the loop, what is the minimum speed the cart must have in order to stay in its circular motion around the loop? (Give your answer to 3 sig figs.)

-Letting the base of the loop represent h=0, how much GPE does the cart have at the top? (Give your answer to 3 sig figs. and in units of J)  

-What is the total mechanical energy of the cart at the top of the loop? (Give your answer to 3 sig figs. and in units of J)

-What speed does the cart have at the base of the loop? (Give your answer to 3 sig figs.)

-What work must be done by the motor to accelerate the cart from rest to its final velocity at the base of the loop? (Give your answer to 3 sig figs. and in units of J. Indicate whether the work is positive or negative with the sign of your answer.)

-When the cart is at the 9 o'clock position of the loop (it has a height above the base of the loop of exactly one loop radius), what is the speed of the cart? (Give your answer to 3 sig figs.)

-When the cart is at the 9 o'clock position of the loop, what is the magnitude of the force of the track exerted on the cart? (Give your answer to 3 sig figs. and in units of N)

-When the cart is at the 9 o'clock position of the loop, what is the magnitude of the torque applied by the force of the track? Let the center of the loop be the axis of rotation. (Give your answer to 3 sig figs. and in units of N*m [which is the SI unit for torque])

-When the cart is at the 9 o'clock position of the loop, what is the magnitude of the net torque exerted on the cart? Let the center of the loop be the axis of rotation. (Give your answer to 3 sig figs. and in units of N*m [which is the SI unit for torque])

Answer 1

consider the force on the cart at the top of the circular loop

F_n = normal force in upward direction

W = weight of the cart in downward direction = m₁ g

V_top = speed of the cart at the top = ?

r₁ = radius of the loop = 17.5 m

force equation for the motion of the cart at the top of loop is given as

W - F_n = m₁ V²_top/ r₁                                            (m₁ V²_top/ r₁   = centripetal force required to move in circle)

for the minimum speed for looping the loop , F_n = 0

hence

W = m₁ V²_top/ r₁     

  m₁ g = m₁ V²_top/ r₁     

V²_top = r₁ g

inserting the values

V²_top = (17.5 x 9.8)

V_top = 13.1 m/s

at the top , H = height = 2 r₁ = 2 x 17.5 = 35 m

gravitational potential energy at the top of loop is given as

GPE = m₁ g H

GPE = (127) (9.8) (35)

GPE = 4.36 x 10⁴ J

total mechanical energy at the top of loop is given as

TE = Kinetic energy at Top + GPE

TE = (0.5) m₁ V_top²+ GPE

TE = (0.5) (127) (13.1)²+ 4.36 x 10⁴

TE = 5.45 x 10⁴ J

V_base = speed of the cart at the base

h = height at the base = 0 m

using conservation of energy

Total energy at the base = total energy at the top of loop

kinetic energy at base + potential energy at base = TE

(0.5) m₁ V_base² + m₁ gh = TE

(0.5) (127) V_base² + (127 x 9.8 x 0) = 5.45 x 10⁴

V_base = 29.3 m/s

W = work done at the base

using work-change in kinetic energy theorem

W = change in kinetic energy

W = (0.5) m₁ V_base²

W = (0.5) (127) (29.3)²

W = 5.45 x 10⁴ J

the work done is positive

h' = height of cart at 9 o clock position = r₁ = 17.5 m

v = speed of cart at 9 o clock position = ?

using conservation of energy between base of the loop and the 9 Oclock position of the cart

Total energy at the base = total energy at 9 oclock position

5.45 x 10⁴ = (0.5) m₁ v² + m₁ g h'

5.45 x 10⁴ = (0.5) (127) v² + (127 x 9.8 x 17.5)

v = 22.7 m/s

F = magnitude of force by the track

force by the track provides the necessary centripetal force , hence

F = m₁ v²/r₁

F = (127) (22.7)² /17.5

F = 3.74 x 10³ N

since the force F is parallel to the radius , hence

Torque = 0 Nm