As shown, a game at a carnival midway involves throwing baseballs at a metal tar

Question

As shown, a game at a carnival midway involves throwing baseballs at a metal target. The target consists of a slender metal rod OB attached to a thin, circular metal disk. (Figure 1) The rod's length is l = 23.0 cm and the disk's radius is r = 7.20 cm . The masses of the rod and disk are 500 and 860 g , respectively. The target is attached to a pin at point O that allows it to swing front to back. What is Itot, the total moment of inertia of the target about point O?

Express your answer numerically to four significant figures with the appropriate units.

Hints

8.837×102 kgm2

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Correct

Part B

As shown, a baseball of mass 145 g is thrown at the target. (Figure 2) The ball hits the target at a vertical distance h = 26.0 cm from the pin at point O with a speed of (vA)1 = 19.0 m/sperpendicular to the target. If the coefficient of restitution between the ball and the target is e = 0.590, what is , the target's angular velocity immediately after the ball hits the target?

Express your answer numerically to three significant figures with the appropriate units.

Hints

11.6 rads

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Part C

If the ball is thrown fast enough at the target, the target will swing into a horizontal position where it will be held in place by a magnet. What is (vA)1min, the minimum velocity of the ball required to swing the target into the horizontal position?

Express your answer numerically to three significant figures with the appropriate units.

Hints

8.837×102 kgm2

Explanation / Answer

moment of inertia of disc that is passing through its center and parallel to its surface along diameter.

= (0.860)(0.072^2)/ 4

= 1.11456 x 10^-3 kg m^2

now applying paralllel axis theorem,

I = Icm + m d^2

I1 = (1.11456 x 10^-3) + (0.860)(0.23 + 0.072)^2

I1 = 0.07955 kg m^2

moment of inertia of rod, I2 = 0.500 x 0.23^2 / 3

= 8.8167 x 10^-3 kg m^2

moment of inertia of system, I = I1 + I2

I = 0.08837 kg m^2

(b) e = velocity of separation / velocity of approach

0.590 = (v2 + wh) / v1

0.590 x 19 = v2 + 0.26w

v2 = 11.21 - 0.26w

now applying angular momentum conservation,

(0.145 x 19 x 0.26) = I w - (0.145 x v2 x 0.26)

0.7163 = 0.08837 w - 0.0377(11.21 - 0.26w)

1.138917 = 0.098172 w

w = 11.6 rad/s ..........Ans

(c) Applying energy conservation to find w.

I w^2 / 2 = m g l / 2 + M g (l + r)

0.08837 w^2 /2 = (0.500 x 9.8 x 0.23 / 2) + 0.860 x 9.8 x (0.23 + 0.072)

0.044185 w^2 = 0.5635 + 2.545256

w = 8.39 rad/s

angular momentum conservation,

(0.145 x v1 x 0.26) = I w - (0.145 x v2 x 0.26)

0.0377 v1 + 0.0377v2 = 0.7414 ..... (i)

and 0.590 = (v2 + wh) / v1

0.590 = (v2 + 2.1814) / v1

0.590 v1 - v2 = 2.1814 ..... (ii)

solving (i) and (ii),

v1 = 13.74 m/s

v2 = 5.92 m/s

Ans: 13.74 m/s

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