Problem 11.10

For this problem use the equation

torque=(moment of inertia) * (angular acceleration).

So you need the moment of inertia I.

You have to use

I=(1/12) ML^2 where M is the mass of the ladder and L is the length.

This is the formula for the moment of inertia of a rod around an axis at its center.

Problem 11.9

Use the formula that

torque = moment of inertia * angular acceleration

Find angular acceleration from the information given: (430 rev/min and 2 revs).

For moment of inertia use the formula for I for a disk.

11.46

For this problem let the axis of rotation be the end of the stick where the paint can is hanging.
Also, assume the whole mass M acts as if at the center of that Mass i.e. 0.5 m away from the
axis of rotation.

Problem 9.15

The impulse is the change in momentum. So compute the vector
J= m*vf - m*vi then compute the angle. Note the angle is pointing towards
the top left quadrant so your angle to the x-axis should be larger than 90.

Problem 9.28

First review your notes. I did a problem very similar to this in class. There
are two steps.

1. First apply conservation of momentum to find the velocity of the putty block system.
2. Then apply conservation of total energy to find the height the putty+block will go.

Problem 9.20

Use conservation of momentum. The initial momentum is zero. Solve for final
momentum of the astronaut. Use this to find velocity and then distance (d=v*t).

Problem 6.28

The key to this problem is to note that the tension T is the same for all segments of the rope.
Once you know this, simply draw the forces acting on the two pulleys. For example the lower pullley will feel a force 2T due to the two segments of rope pulling it UP, it will also feel a
force mg pulling it down. The net force will yield the acceleration of the box.