- #1

- 26

- 0

## Homework Statement

A 2.3 kg wood block hangs from the bottom of a 1.3 kg, 1.3 meter long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 35 degree angle.

## Homework Equations

Conservation of angular momentum

Conservation of energy

Moment of Inertia

## The Attempt at a Solution

m

_{b}= .012 kg

m

_{B}= 2.3 kg

m

_{R}= 1.3 kg

L = 1.3 m

r = .65m

v

_{i}= ?

-Conservation of Angular momentum:

A

_{i}= A

_{f}

A

_{i}= (m

_{b})(v

_{i})(L) = I

_{T}(w) = A

_{f}

Where w is the final angular velocity

Where I

_{T}is the total moment of inertia of the system. Given by:

(1/3)((m

_{R})(r)

^{2}+ (m

_{b}+ m

_{B})(L

^{2})

So,

1: w =([m

_{b}* v

_{i}* L)/I

_{T}

-Conservation of Energy

KE = Change in PE

KE = (1/2)(I

_{T})w

_{2}

Change in PE, treating initial position of the pendulum as PE = 0:

PE = m

_{T}* g * h

m

_{T}is the sum of the masses

Where h is the change in height, denoted by the change in the center of mass as the pendulum rotates:

center of mass = c = (m

_{R}* L + (mB + mb)*r)/mT

h = c - c*cos(35)

Equating KE and PE, then solving for w:

w

_{2}= (m

_{T}* g * h)*2/I

_{T}

So w is the square root of all that mess.

Equating the above equation with 1 and solving for v

_{i}

v

_{i}= (I

_{T}* sqrt{ (m

_{T}* g * h)*2 / I

_{T}}) / ((m

_{b})(L))

Which gives me something like 441. I'm really frustrated with this and I'm not sure what I did wrong. Thanks in advance.