Conservation of Angular Momentum and Uniform Rod

[peaceandlove]

Homework Statement

A small 0.157 kg block slides down a frictionless surface through height h = 0.307 m and then sticks to a uniform vertical rod of mass M = 0.397 kg and length d = 2.78 m. The rod pivots about point O through angle θ before momentarily stopping. Find θ.

Homework Equations

Equation 1: I= (1/3)Md^2 + md^2
Equation 2: v=(2gh)^(1/2)
Equation 3: W=m(block)dv/I
Equation 4: arccos(1 -(I*W^2/[d*g(2m + M) ]))

[I and W are for the road and block system]

The Attempt at a Solution

Using equation 1, I found I to be 0.02727 kg*m^2. Then I found W to be 4.3357 using equation 2 and 3. I then plugged in the values into equation 4 and got 40.50412 degrees, but apparently I'm doing something wrong.

 

[peaceandlove]

Nevermind!

 

[nlightner]

I would first like to commend the student for their attempt at solving the problem and using the appropriate equations. However, there seems to be a mistake in the calculation of I and W.

Equation 1, which is the moment of inertia formula for a rod, should be I= (1/12)Md^2 + md^2. Using this formula, I= 0.00162 kg*m^2.

Equation 2, which gives the velocity of the block at the bottom of the ramp, should be v= (2gh)^(1/2). Using this formula, v= 1.235 m/s.

Equation 3, which gives the angular velocity of the rod and block system, should be W= m(block)*v/I. Using this formula, W= 758.64 rad/s.

Finally, plugging in these values into equation 4, we get θ= 52.334 degrees. This is the correct answer for the angle through which the rod pivots before stopping.

It is important to carefully check the units and equations used in a problem, as a small mistake can lead to incorrect results. In addition, it is always helpful to double check the calculations and make sure they make sense in the context of the problem. Good job on attempting to solve the problem and keep up the good work!