Solving a Uniform Rod's Angular and Linear Speed

[postfan]

Homework Statement

A uniform rod of mass 1.2 kg and length 1.2 m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released. The acceleration due to gravity is g = 9.8 m/s2.

What is the angular speed (in rad/s, but do not include units) of the rod as it passes through the vertical position (when end marked B is at the bottom)?

What is the linear speed of the bottom end of the rod (marked B) when it passes the vertical position?(in m/s, but do not include units)

Homework Equations

The Attempt at a Solution

Used conservation of energy, m*g*h=.5*I*w^2. I calculated I to be .252 and used .3 for h. Solving for w, I got 4.83. Then used the formula v=wr and got v=2.898. What am I doing wrong?

 

[gneill]

Homework Statement

A uniform rod of mass 1.2 kg and length 1.2 m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released. The acceleration due to gravity is g = 9.8 m/s2.

What is the angular speed (in rad/s, but do not include units) of the rod as it passes through the vertical position (when end marked B is at the bottom)?

What is the linear speed of the bottom end of the rod (marked B) when it passes the vertical position?(in m/s, but do not include units)

Homework Equations

The Attempt at a Solution

Used conservation of energy, m*g*h=.5*I*w^2. I calculated I to be .252 and used .3 for h. Solving for w, I got 4.83. Then used the formula v=wr and got v=2.898. What am I doing wrong?

Your method looks fine. The numerical value of your moment of inertia looks okay, as does your value for the Δh of the center of mass. Still, something went wrong when you calculated the value of ω. Better check your math there.

 

[postfan]

I found that w=5.2915, and v=3.1749. Is that right?

 

[gneill]

I found that w=5.2915, and v=3.1749. Is that right?

ω looks better. What did you use for the radius of arc for the bottom end of the rod?

 

[postfan]

I used .6 (length of rod /2)

 

[ehild]

I used .6 (length of rod /2)

But B is at 3/4 L distance from the pivot...


ehild

 

[postfan]

Ok is the radius then .45 (length of rod*.75/2)?

 

[ehild]

Ok is the radius then .45 (length of rod*.75/2)?

Why do you divide by 2?


ehild

 

[postfan]

Radius is 1/2 the diameter. The diameter is .9.

 

[gneill]

Radius is 1/2 the diameter. The diameter is .9.

You might want to rethink that. The radius extends from the center of the motion (the pivot) out to the point in question (point B).

 

[postfan]

OK, so the radius is .9 and the linear velocity is 4.76235, right?

 

[gneill]

OK, so is the radius .9?

Are you guessing and looking for confirmation? It would be preferable to describe your logic and conclusion. Why .9m?

Hint:
Why don't you draw a diagram with the rod in the initial and final positions, and sketch in the arc that B describes. Where's the center of the arc? What's its radius?

 

[gneill]

OK, so the radius is .9 and the linear velocity is 4.76235, right?

That looks better. Now, reconcile the significant figures in your result with the given data (5 decimal places is certainly not justified )

 

[postfan]

Ok, so is it just 4.8?

 

[gneill]

Ok, so is it just 4.8?

4.8 looks fine numerically, but it would be marked WRONG if you don't include the units