A thin, uniform, 3.80 kg bar, 80.0 cm long, has very small 2.50 kg balls glued on at either end. It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right-hand ball becomes detached and falls off, but the other ball remains glued to the bar.
(a) Find the angular acceleration of the bar just after the ball falls off.
(b) Will the angular acceleration remain constant as the bar continues to swing?

Net force acting on the rod which will try to rotate will be weight of 2.5 kg ball . Because it was this force which was balancing the whole system. Torque due to this force = mg x r

= 2.5 x 9.8 x .4 Nm

= 9.8 Nm

angular acceleration = torque / moment of inertia

= 9.8 / .60267

= 16.26 rad / s²

b ) This angular acceleration will change because torque will change ( decrease in fact ) . It is to be noted that maximum value of torque acts when force is perpendicular . Since torque decreases , angular acceleration decreases.

snehashish65 snehashish65 · Answer 2 · 2021-11-18T09:50:38+01:00

(a) The angular acceleration of the bar just after the ball falls off is [tex]16.26 \;\rm rad/s^{2}[/tex].

(b) The angular acceleration will not remain constant due to variation in torque.

Given data:

The mass of bar is, m = 3.80 kg.

The length of bar is, L = 80.0 cm = 0.8 m.

The mass of balls is, m' = 2.50 kg.

(a)

First we need to calculate the moment of inertia such that a ball is attached at one if its ends;

Then required moment of inertia is,

[tex]I = \dfrac{ mL^{2}}{12}+ \dfrac{m'L}{2}\\\\I = \dfrac{ 3.80 \times 0.8^{2}}{12}+ \dfrac{2.50 \times 0.8}{2}\\\\I=0.60267 \;\rm kgm^{2}[/tex]

Now, the net force acting on the rod which will try to rotate will be weight of 2.5 kg ball . Because it was this force which was balancing the whole system. So, torque is,

[tex]T = W \times \dfrac{L}{2} \\\\T = m'g \times \dfrac{L}{2} \\\\T = 2.5 \times 9.8 \times \dfrac{0.8}{2}\\\\T = 9.8 \;\rm Nm[/tex]

So, angular acceleration is,

[tex]\alpha=\dfrac{T}{I}\\\\\alpha=\dfrac{9.8}{0.60267} = 16.26 \;\rm rad/s^{2}[/tex]

Thus, the angular acceleration of the bar just after the ball falls off is [tex]16.26 \;\rm rad/s^{2}[/tex].

(b)

No, the angular acceleration will not remain constant because torque will change ( decrease in fact ) . It is to be noted that maximum value of torque acts when force is perpendicular . Since torque decreases , angular acceleration decreases.

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