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Salsk061 [2.6K]
3 years ago
11

You throw a bowling ball down an alley with speed v_0v ​0 ​​ , and initially it slides without rolling. After a little while fri

ction kicks in and the ball begins to roll. What is the speed of the bowling ball when it starts rolling without slipping?
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

final speed after pure rolling is given as R\omega = \frac{5}{7}v_0

Explanation:

As we know that point of contact at ground is taken as reference then there is no external torque about this point on the ball

So we can use angular momentum conservation about this point

L_i = L_f

L_i = mv_0R

L_f = \frac{7}{5}mR^2\omega

so we have

mv_0R = \frac{7}{5}mR^2\omega

R\omega = \frac{5}{7}v_0

So final speed after pure rolling is given as R\omega = \frac{5}{7}v_0

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The distance traveled by the particle at the given time interval is 0.28 m.

<h3>Position of the particle at time, t = 0</h3>

The position of the particle at the given time is calculated as follows;

x = 2 sin2(t)

y = 2 cos2(t)

x(0) = 2 sin2(0) = 0

y(0) = 2 cos2(0) = 2(1) = 2

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1 year ago
A car travels 100 m while decelerating to 8 m/s in 5 s.<br> a) What was its initial speed?
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Answer:

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

v_{f} =v_{i} - (a*t)\\

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

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Vi = initial velocity [m/s]

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Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

v_{f}^{2} = v_{i}^{2} - (2*a*d)

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d = distance = 100[m]

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0 = 80*a - 200*a + 25*a^2

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8 = Vi - (4.8*5)

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