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Hi I am kind of stuck on this question.

Ok here is my attempt to the solution:

I did my derivation of the moment of inertia no problem. and using the parallel axis theorem, the new moment of inertia is I=Mx^"+Io where Io is the moment of inertia through the centre of mass of the rod.

a) the centre of mass of the rod will move forward and the rod will also rotate about the new centre of mass which is a distance x away from the original i.e. at l/2

b)Using conservation of momentum

mv=(m+M)v1 which gives 1/2Mv=(3/2)Mv1

so v1=1/3v

Part C is where I am begining to get stuck.

c) using conservation of energy and angular momentum

[tex]\frac{1}{2}(\frac{1}{2}M)v^2=\frac{3}{2}M(\frac{1}{3}v)^2+\frac{1}{2}I\omega^2[/tex]

this gives [tex]\frac{1}{6}Mv^2=\frac{1}{2}I\omega^2[/tex]

from angular momentum

1/2Mv(l/2-x)=I*omega where x is the distance between th new and old centre of mass

from this Mv =(2*I*omega)/(l/2-x) and substitute this into the previous equation gives 1/6*(2*I*omega)v/(l/2-x)=1/2I*omega^2

this gives omega=2/3 v/(l/2-x)

d) substitute omega=2/3 v/(l/2-x)back into 1/2Mv(l/2-x)=I*omega gives

I=3/4M(l/2-x)^". this is the moment of inertia of rod and poin mass combined about the axis which passes through the new centre of mass

the moment of inertia Io of combined masses through the initial centre of mass is Io=Ml^2/12+1/2m(l/2)^2=Ml^2/12+Ml^2/16=7Ml^2/48

using parralel axis theorem Io=Mx^2+I

solving the quadratic equation gives x=(3+/-sqrt(3))l/18

I really don't know if I am on the right track or not , could somebody help me please. thanks

**Show that the moment of inertia of a uniform thin rod of mass M and length l about an axis perpendicular to the rod through its centre is**

[tex]\frac{Ml^2}{12}[/tex]

What is the moment of inertia of the rod about a parallel axis a distance x from the centre?

The rod defined above lies at rest on a smooth horizontal table. A point

mass m = 1/2M moving at right angles to the rod with a speed v collides with one end of the rod, and sticks to it.

(a) Describe qualitatively, including diagrams, the subsequent motion of the

rod.

(b) Derive the speed and direction of motion of the centre of mass (of the

rod and mass combined), after the collision.

(c) Derive the angular speed of the combined rod and mass after the

collision.

(d) Find the position on the rod that is stationary immediately after the

collision. [[tex]\frac{Ml^2}{12}[/tex]

What is the moment of inertia of the rod about a parallel axis a distance x from the centre?

The rod defined above lies at rest on a smooth horizontal table. A point

mass m = 1/2M moving at right angles to the rod with a speed v collides with one end of the rod, and sticks to it.

(a) Describe qualitatively, including diagrams, the subsequent motion of the

rod.

(b) Derive the speed and direction of motion of the centre of mass (of the

rod and mass combined), after the collision.

(c) Derive the angular speed of the combined rod and mass after the

collision.

(d) Find the position on the rod that is stationary immediately after the

collision. [

Ok here is my attempt to the solution:

I did my derivation of the moment of inertia no problem. and using the parallel axis theorem, the new moment of inertia is I=Mx^"+Io where Io is the moment of inertia through the centre of mass of the rod.

a) the centre of mass of the rod will move forward and the rod will also rotate about the new centre of mass which is a distance x away from the original i.e. at l/2

b)Using conservation of momentum

mv=(m+M)v1 which gives 1/2Mv=(3/2)Mv1

so v1=1/3v

Part C is where I am begining to get stuck.

c) using conservation of energy and angular momentum

[tex]\frac{1}{2}(\frac{1}{2}M)v^2=\frac{3}{2}M(\frac{1}{3}v)^2+\frac{1}{2}I\omega^2[/tex]

this gives [tex]\frac{1}{6}Mv^2=\frac{1}{2}I\omega^2[/tex]

from angular momentum

1/2Mv(l/2-x)=I*omega where x is the distance between th new and old centre of mass

from this Mv =(2*I*omega)/(l/2-x) and substitute this into the previous equation gives 1/6*(2*I*omega)v/(l/2-x)=1/2I*omega^2

this gives omega=2/3 v/(l/2-x)

d) substitute omega=2/3 v/(l/2-x)back into 1/2Mv(l/2-x)=I*omega gives

I=3/4M(l/2-x)^". this is the moment of inertia of rod and poin mass combined about the axis which passes through the new centre of mass

the moment of inertia Io of combined masses through the initial centre of mass is Io=Ml^2/12+1/2m(l/2)^2=Ml^2/12+Ml^2/16=7Ml^2/48

using parralel axis theorem Io=Mx^2+I

solving the quadratic equation gives x=(3+/-sqrt(3))l/18

I really don't know if I am on the right track or not , could somebody help me please. thanks

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