(o) obtained from (b). 3. Ball A is attached to one end of a rigid massless rod,

Question

(o) obtained from (b). 3. Ball A is attached to one end of a rigid massless rod, while an identical ball B is attached to the center of the rod, as the right figure illustrates. Each ball has a mass of m 0.50 kg, and the length of each half of the rod is L 0.40 m. This arrangement is held by the empty end and is whirled around in a horizontal circle at a constant rate, so each ball is in uniform circular motion. Ball A travels at a constant speed of VA 5.0 m/s. (14 4+4+6 points) a. Calculate the speed of ball B, vB. b. Obtain the magnitude of the accelerations of both balls Find the tension in each half of the rod

Explanation / Answer

Given that,

mA = mB = 0.5 kg ; L = 0.4 m, vA = 5 m/s

(a)we need to find the speed of the ball B, vB

We see from the figure that, the ball A is covering distance equal to the circumference of the circle that is

2 pi R = 2 pi (2L) while B is travellling only 2 pi L in the same time as that of ball A.

So the speed of the ball B must be half of the speed of ball a.

Hence, vB = 5/2 = 2.5 m/s

(b)We know that, F = mv2/r = ma => a = v2/r

The acceleration of ball A will be :

aA = (vA)2/2L = (5)2/2x0.4 = 25/0.8 = 31.25 m/s2

The acceleration of ball B will be :

aB = (vB)2/L = (2.5)2/0.4 = 6.25/0.4 = 15.63  m/s2

Hence, aA = 31.25  m/s2 and aB = 15.63  m/s2

(c)Let, TA be the tension in the rod between the two balls and it provides the required centripital force for keeping the ball A on its circular path.

TA = m(vA)2/2L = 0.5 x (5)2 / 2 x 0.4 = 15.63 N

At ball B, TB - TA = m(vB)2/L => TB = m(vB)2/L + TA

TB = 0.5 x (2.5)2 / 0.4 + 15.63 = 7.82+15.63 = 23.45 N

Hence, TA = 15.63N and TB = 23.45N.

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