- #1

- 3

- 0

## Homework Statement

There is pulley which has a mass hanging from one side, and a rotating rod attached horizontally to the other (both by cables). The pulley has mass; M

_{p}= 9.5kg and radius; R

_{p}=0.2m. The block(A) has mass; M

_{a}=10.9kg. The rod is L = 0.8m long and has mass=M

_{l}7.9kg. The rod is also rotating at w= 2.7 rad/s in the anti clockwise direction. The value for gravity used is g= 9.8m/s/s downwards. Tensions are denoted by T

_{a}and T

_{b}

## Homework Equations

Newtons 2nd Law(F=ma), Eulers Equation (M=I[itex]\alpha[/itex]).

## The Attempt at a Solution

I have defined my coordinate system as anticlockwise is positive and vertically down is positive.

I

_{pulley}= 0.5MR

^{2}

I

_{rod at centre of grav}=(1/12)ML

^{2}

I have developed some simultaneous equations for the system:

T

_{a}- M

_{a}g = M

_{a}a

_{a}From Newton 2nd Law of Block(A)

a

_{a}= R

_{p}[itex]\alpha[/itex]

_{p}Acceleration relationship

-T

_{a}R

_{p}+T

_{b}R

_{p}=I

_{p}[itex]\alpha[/itex]

_{p}Eulers equation for pulley

T

_{b}- M

_{r}g = M

_{r}a

_{r}Newtons second law on rod.

I also have the following(which i think is where the problem is at):

-T

_{b}L - I

_{r}[itex]\alpha[/itex]

_{r}= -Lw

^{2}Eulers equation on the rod at the centre of gravity.

a

_{r}= -L[itex]\alpha[/itex]

_{r}

I solved this system, and got an incorrect answer. As above i think the problem lies in the last two equations.

Last edited: