How long will it take for a basketball spinning on someone’s finger to stop if it undergoes and angular acceleration of -0.15 rad/s and is traveling at 27 rad/s initially? How many revolutions will it make?

Answers

Answer 1

The basketball's angular velocity ω at time t is

ω = 27 rad/s + (-0.15 rad/s²) t

It comes to a stop when ω = 0, which happens for

0 = 27 rad/s - (0.15 rad/s²) t

t = (27 rad/s) / (0.15 rad/s²)

t = 180 s

In this time, the ball would undergoes an angular displacement θ of

θ = (27 rad/s) t + 1/2 (-0.15 rad/s²) t ²

Plug in t = 180 s and solve for θ :

θ = (27 rad/s) (180 s) - 1/2 (0.15 rad/s²) (180 s)² = 2430 rad

One complete revolution corresponds to a turn of 2π rad, so the ball makes

(2430 rad) / (2π rad/rev) ≈ 386.747 rev

or about 387 revolutions as it slows down.


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A 0.144 kg baseball approaches a batter with a speed of . The batter lines the ball directly back to the pitcher with a speed of . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat 20 m/s 30 m/s.

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Helppp!!!!! Please!!
A 0.144 kg baseball approaches a batter with a speed of . The batter lines the ball directly back to the pitcher with a speed of . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat 20 m/s 30 m/s

Answers

Complete question:

A 0.144 kg baseball approaches a batter with a speed of 20m/s . The batter lines the ball directly back to the pitcher with a speed of 30m/s . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat

Answer:

(1) The impulse exerted by the ball is 7.2 kg.m/s

(2) The average force exerted on the ball by the bat is 600 N.

Explanation:

Given;

mass of the baseball, m = 0.144 kg

velocity of the baseball, v₁ = 20 m/s

velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)

The impulse exerted by the ball is calculated as follows;

J = ΔP = mv₁ - mv₂

ΔP = m(v₁ - v₂)

ΔP = 0.144 [20 - (-30)]

ΔP = 0.144 ( 20 + 30 )

ΔP = 0.144 (50)

ΔP = 7.2 kg.m/s

The average force exerted on the ball by the bat is calculated as;

[tex]F = \frac{\Delta P }{t} \\\\where; t \ is \ time \ of \ contact= 0.012 \ s\\\\F = \frac{7.2}{0.012} \\\\F = 600 \ N[/tex]

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