Physics, 20.03.2020 01:22 TPfishing16

A uniform thin rod of length 0.84 m and mass 4.6 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.9 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the bullet's speed just before impact?

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Answer from: jesse7412

The velocity of the bullet is v_o=2.2886 *10^4 m/s

Explanation:

The free body diagram of this question is shown on the first uploaded image

Since we are told that the bullet got lodged in the rod we are going to look at the bullet and rod combined as a system

And this also means that the collision is inelastic since the both object stuck together

examining this question we can deduce that the pivot on the rotational axis also exerts an external force on the system (the rod + the bullet)

These eternal force does not cause rotation but hold the rod along the horizontal plane hence the these is no external which mean that the angular momentum is conserved

Generally angular momentum is mathematically represented as

L_b = +mvb

This positive because the bullet misses the axis to the right (Which means it would be moving in a counter clock- wise direction around the axis)

Before the collision v =v_o

And b is the minimum distance of approach as shown on the diagram

According to trigonometry

$b = \frac{l}{2} sin \theta$

Now the total angular momentum before collision is

$L_i = L_b = m v_o\frac{l}{2} sin(60^o)$

Generally moment of inertia is mathematically represented as

I = mr^2

Where m is the mass and r is the distance to the axis of rotation

After collision since the bullet and the rod combined and became a system

their moment of inertia would be the same and is mathematically given as

Assuming that it was $\frac{1}{3} \ of \ Mass \ of\ the \ rod \ that was\ hit\ by \ the \ bullet \ since \ they \\ stuck \ together$

$I = I_{rod} + mr^2 = \frac{1}{3} M(\frac{l}{2}) ^2 +m(\frac{l}{2} )^2 = (\frac{M}{12} +\frac{m}{4} )l^2 = \frac{1}{12}(M + 3m)l^2$

$I = I_{rod} + mr^2 = \frac{1}{3 *4} Ml^2 +m(\frac{l}{2} )^2 = (\frac{M}{12} +\frac{m}{4} )l^2 = \frac{1}{12}(M + 3m)l^2$

$I = I_{rod} + mr^2 = \frac{1}{12} Ml^2 +m(\frac{l}{2} )^2 = (\frac{M}{12} +\frac{m}{4} )l^2 = \frac{1}{12}(M + 3m)l^2$

Now as the system rotates due the force of the bullet and the fact that the plane is fixed about the vertical axis at an angular speed of

w_f = 12.0 rad/s

The final momentum is mathematically represented as

$L_f = Iw_f = \frac{1}{12}(M +3m)l^3w_f$

Since Angular momentum is conserved

L_i = L_f

$mv_o\frac{l}{2}sin(60^o) = \frac{1}{12} (M +3m)l^2w_f$

Making v_o the subject of the formula gives

$v_o = \frac{2}{mlsin(60^o)} \cdot \frac{1}{12} (M+3m)l^2w_f =\frac{(M+3m)lw_f}{6msin(60^o)}$

$= \frac{(4.6Kg+(3* 3.9 *\frac{1kg}{1000} ))(0.84m)(12 rad/s)}{6(3.9*\frac{1Kg}{1000} )(\frac{\sqrt{3} }{2} )}$

v_o=2.2886 *10^4 m/s

A uniform thin rod of length 0.84 m and mass 4.6 kg can rotate in a horizontal plane about a vertica

Answer from: Quest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

Answer from: Quest

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