A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball (the f

Question

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball (the figure) slides on the lane with initial speed vcom,0 = 8.5 m/s and initial angular speed omega0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.33. The kinetic frictional force fk acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed vcom has decreased enough and angular speed psi has increased enough, the ball stops sliding and then rolls smoothly. During the sliding, what are the ball's linear acceleration and angular acceleration? How long does the ball slide? How far does the ball slide? What is the linear speed of the ball when smooth rolling begins? Note that the clockwise direction is taken as negative.

Explanation / Answer

radius of the ball R=11cm

Vcm=8.5 m/sec

kinetic friction coefficient uk=0.33

a)

linear acceleration a=uk*g

a=0.33*9.8

a=3.234 m/sec2

b)

here,

frictional force fK=uk*(mg)

and

torque =R*fK

I*alpa=R*fk

(2/5 *m*R^2)*alpa=R*(uk*mg)

(2/5)R*alpa=(uk*g)

alpa=(5/2)*(uk*g)/R

alpa=(5/2)*(0.33*9.8)/(0.11)

angular acceleration, alpa=73.5 rad/sec^2

c)

Vcm=Vo+a*t

and

W=alpa*t

but,

Vcm=R*w

(vo+a*t)=R*(alpa*t)

t=vo/(R*alpa-a)

t=8.5/(0.11*73.5-(-3.234))

t=0.751 sec

d)

distance travelled,

x=vo*t+1/2*a*t^2

x=(8.5*0.751)-(1/2)*(3.234)*(0.751^2)

x=5.471 m

e)

rolling speed (at begining) is,

v=vo+a*t

v=8.5-(3.234*0.751)

v=6.07 m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.