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hw8solution

# hw8solution - Homework 8 Solution Y&F 7.46(a From the...

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Homework 8 Solution Y&F 7.46 (a) From the discussion in class we know that for the car to go around the loop without falling off the track, the required minimum velocity at the top of the loop (point B in the figure) has to be B gR . Set up the energy equation and ignore the elastic potential energy (no spring) and W 1 2 (no friction): K A U gA K B U gB Take the ground level as the datum and with A 0 (released from rest) we have 1 2 m A 2 mgy A 1 2 m B 2 mgy B mgh min 1 2 m B 2 mgy B gh min 1 2 gR 2 g 2 R h min 2.5 R (b) We use the energy equation again for the motion from point A to point C to find the velocity at point C : K A U gA K C U gC 1 2 m A 2 mgy A 1 2 m C 2 mgy C mgh 1 2 m C 2 mgy C mg 3.5 R 1 2 m C 2 mg R C 5 gR 5 9.8 20 31.3 m/s With the velocity at point C calculated, the normal acceleration can now be found to be a n C 2 R 5 gR R 5 g 5 9.8 49 m/s 2 We apply Newton’s 2nd law to find the tangential acceleration. At point C the tangential axis is pointing downward (car is moving vertically down), and the only force in the vertical direction is the weight of the car. Therefore, F t ma t mg ma t a t g 9.8 m/s 2 CHANG 1 (a) The minimum velocity at the top of the loop can be found by making tension in the wire zero at the top of the loop, and with the weight as the only force producing centripetal acceleration, we have from Newton’s 2nd law mg m 2 / R gR at the top. This is the same result that we had for a particle completing a vertical loop along the circular track. Therefore the general conclusion is that for a particle to complete a vertical circular loop in any fashion, the minimum required velocity has to be gR at the top of the loop. So for the sphere to complete the loop, its speed at the top of the circle has to have a minimum value of gR gL (the radius of the loop equals the length of the string). Label positions 1 and 2 for the starting point and the top of the loop respectively. Taking the center

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