(a) A ball of mass 0.130 kg is dropped from rest from aheight of 1.25 m. It rebo
ID: 1681535 • Letter: #
Question
(a) A ball of mass 0.130 kg is dropped from rest from aheight of 1.25 m. It rebounds from the floor to reach a height of0.920 m. What impulse was given to the ball by the floor?____________ kg·m/s (b) A potter's wheel -- a thick stone disk of radius 0.500 mand mass 105 kg -- is freely rotating at 50.0 rev/min. The pottercan stop the wheel in 6.00 s by pressing a wet rag against the rimand exerting a radially inward force of 65.0 N. Find the effectivecoefficient of kinetic friction between the wheel and therag. _____________ (i got .063, but that was the wrong answer) (a) A ball of mass 0.130 kg is dropped from rest from aheight of 1.25 m. It rebounds from the floor to reach a height of0.920 m. What impulse was given to the ball by the floor?
____________ kg·m/s (b) A potter's wheel -- a thick stone disk of radius 0.500 mand mass 105 kg -- is freely rotating at 50.0 rev/min. The pottercan stop the wheel in 6.00 s by pressing a wet rag against the rimand exerting a radially inward force of 65.0 N. Find the effectivecoefficient of kinetic friction between the wheel and therag. _____________ (i got .063, but that was the wrong answer)
Explanation / Answer
mass m = 0.13 kg Initial height h = 1.25 m final height h ' = 0.92 m Initial speed when it strikes the ground u =[ 2gh ] = 4.949 m / s final speed when it leavs the ground v = - [ 2gh' ] =- 4.246 m / s impilse P = m ( u-v ) = 1.195 kg m / s (b). Radius r= 0.5 m mass m = 105 kg Initial anglar speed w = 50 rev / min = 50 * 2 rad / 60 s = 5.235 rad / s time t = 6 s final angular speed w ' = 0 Angular accleration = ( w' - w) / t= -0.872 rad / s^2 torque T = I where I = moment of inertia = ( 1/ 2) m r ^ 2= 13.125 kg m^2 So, T = 11.445 N m force F = 65 N the effective coefficient of kinetic friction between thewheel and the rag = ? we know = -g frm this = 0.0089Related Questions
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