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e-lub [12.9K]
3 years ago
15

What is the solution for the gravitational pull on earth?

Physics
1 answer:
chubhunter [2.5K]3 years ago
8 0

The magnitude of the force of gravity pulling on an object located on Earth is

   (9.8 meters per second²) x (the mass of the object in kilograms).

The direction of the force is toward the center of the Earth.

The unit for the magnitude of the force is [ Newton ].

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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. T
nasty-shy [4]

Answer:

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

                           

                               

                     

               

                 

8 0
3 years ago
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