A spherical bowling Dali with mass m = 4. 1 kg and radius R - 0.117 m is thrown

Question

A spherical bowling Dali with mass m = 4. 1 kg and radius R - 0.117 m is thrown down the lane with an initial speed of v - 8.9 m/s. The coefficient of kinetic friction between the sliding bail and the ground is mu = 0.25. Once the ball begins to roll without slipping it moves with a constant velocity down the lane. What is the magnitude of the angular acceleration of the bowling bail as it slides down the lane? rad/s^2 What is magnitude of the linear acceleration of the bowling Dali as it slides down the lane? m/s^2 How long does it take the bowling bail to begin rolling without slipping? How far does the bowling bail slide before It begins to roll without slipping? s What is the magnitude of the final velocity? m/s After the bowling ball begins to roll without slipping. Compare the rotational and translational kinetic energy of the bowling ball: KE_rot KE_tran

Explanation / Answer

According to the given problem,

First , to calculate angular acceleration we need liner accelartion,so calculate Q.2

2)Liner acceeration is,

a = µg = 0.25 * 9.81m/s²

a = 2.4525 m/s² 2.45 m/s²

Now, calulate the angular acceleration

1) Angular acceleration is,

= a / r = 2.45m/s² / 0.117m

= 20.94 rad/s² 21.0 rad/s²

3) As it begins to rolling when,

= V / r.
= *t = 21 rad/s² * t we need to calulate the V

and V = Vo - a*t = 8.9m/s - 2.45m/s² * t

Then

21 rad/s² * t = (8.9m/s - 2.45m/s²*t) / 0.117m

multiply by 0.117m and drop units for ease (t is in seconds)

2.457 * t = 8.9 - 2.45*t
t = 8.9 / 4.907

t = 1.813 s 1.8 s

4) Now, the distance travelled is ,

s = Vo*t - ½at² =

s= (8.9*1.8 - ½*2.45*(1.8)²) m

s = 12.051 m 12 m

5) The final velocity is,

V = Vo - at = 8.9m/s - 2.45m/s² * 1.8 s

V = 4.49 m/s 4.5 m/s

6) First, calcualte the traslational K.E

trans K.E = ½ * 4.1kg * (4.5m/s)²

= 41.5125J 41.5J

Then calculate rotational K.E,

I = (2/5)mr² = (2/5) * 4.1kg * (0.117m)² = 0.02245 kg·m²

= v/r = 4.5m/s / 0.117m = 38.46 rad/s,

so rot KE = ½I²

= ½ * 0.02245 kg·m² * (38.46rad/s)²

= 11.05 J

KErot < KEtran.

I think you understand my explanation, and hope you got your solution and glad I could help. Rate me! if you like the solution.

Thank you.

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