Problem 3. A uniform bowling ball of radius R, mass M is thrown down a horizonta

Question

Problem 3. A uniform bowling ball of radius R, mass M is thrown down a horizontal lane ithita horizontal speed vo and backspintas shown beow) with initial angular speed o, such that vo RDo. So after the ball makes contact with the horizontal lane, t rolls with slipping on the lane. The kinetic frictional force f acting on the ball causes a center of mass acceleration of the bal acom to slow down its linear motion. Meanwhile this kinetic frictional force f aso produces a torque that causes an angular acceleration of the ball to speed up its angular motion. When speed Va has decreased enough and the angular speed has ncreased enough, the ball wil start to roll without slipping. The coefficient of kinetic friction between the ball and lane surfaces is k. The moment of merta of the ball about its center of mass is loom-2SMR (a) Find the center of mass velocity v and angular velocity ofthe ball as afunction of time after it makes contact with the lane and up to the point when it starts to roll without slipping. Note: directions are important her! (b) Find the center ofmass velocity vr and angular velocity without slipping (c) Find the frictional force between the bowling ball and horizontal surface during the ball's rolling without slipping phase; d) Find the center of mass acceleration acom of the ball during its rolling without slipping phase e) Find the angular acceleration of the ball during its rolling without slipping phase; ) If the thrower of the ball wants the ball to come to rest before it can reach the point that it will art to roll without slipping, how must the initial conditions, vo and 0o,be related? of the bal when t just starts to rol Wf Vo

Explanation / Answer

given mass = M

radius = R

intiial horizontal sped = vo

initial angular speed = wo

fricintoal force = fk

acceleration of com = ac

coefficient of friciont = k

I = 2MR^2/5

a. beofre the ball starts to roll without slipping

after time t

fk = k*Mg

hence ac = k*g

hence

v = vo - kg*t

and

w = wo + alpha*t

alpha = 5k*M*g*R/2MR^2 = 5kg/2R

hence

w = wo + 5kgt/2R

b. when the ball starts to roll without slipping

v = wR

vo - kgt = woR + 5kgt/2

t = 2(vo - woR)/7kg

hence

v = vo - kg*2(Vo - woR)/7kg

v = vo - 2Vo/7 + 2woR/7

v = 5Vo/7 + 2woR/7

and w = v/R = 5Vo/7R + 2wo/7

c. for rolling without slipping

linear acceleration = a

angular accleration = a/R

now, friction force = f

f = M*a

and

f*R = 2MR^2*a/(5R)

f = 2Ma/5

hecne a = 0m/s/s

d. hence f = 0 N

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