What Is the Speed of a Sledgehammer's Center of Mass After Being Released?

[Becca93]

Homework Statement [/b]
A sledgehammer with a mass of 2.30 kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_cm=0.570 m, and the moment of inertia about the center of mass is I_cm = 0.0390 kgm2. If the hammer is released from rest at an angle of theta=40.0 degrees such that H=0.366 m, what is the speed of the center of mass when it passes through horizontal?

The diagram is attached.

The Attempt at a Solution

My prof did a similar question during class with energy, and so that's the way I tried to solve it. With this incorrect, I honestly don't know how else to go about this problem.

E1 = E2
mgH = (1/2)Iw^2
2mgH/I = w^2
w = √(2mgH/I)
w = 20.568 rad/s

*Because the question asks for speed, I tried this as the answer and the units were rejected. So I tried to solve for velocity instead.

mgH = (1/2)I(v^2/r^2)
v^2 = 2mgHr^2/I
v = √(2mgHr^2/I)
v = 19.37 m/s

Again, incorrect.

Does anyone know how to help? I would genuinely appreciate it. I've got a final coming up and not knowing how to go about this problem, or even what specifically to solve for when asked for 'speed', is really making me panic.

 

[Doc Al]

E1 = E2
mgH = (1/2)Iw^2
2mgH/I = w^2
w = √(2mgH/I)
w = 20.568 rad/s

So far, so good. You found the angular speed. Now use it to answer the question: What's the speed of the center of mass? (They want the linear speed.)

 

[Becca93]

So far, so good. You found the angular speed. Now use it to answer the question: What's the speed of the center of mass? (They want the linear speed.)

v = wr
so, with the angular speed I have,
v = (20.568)(0.570)
v = 36.1 m/s
Which is incorrect.

Where am I going wrong?

 

[Doc Al]

v = wr
so, with the angular speed I have,
v = (20.568)(0.570)
v = 36.1 m/s
Which is incorrect.

Where am I going wrong?

Check your arithmetic.

 

[Becca93]

Check your arithmetic.

Oh, okay.
But I end up with 11.7, which is still the incorrect answer.

 

[Doc Al]

The distance from the pivot to the center of mass is r_cm=0.570 m, and the moment of inertia about the center of mass is I_cm = 0.0390 kgm2.

They give the moment of inertia about the center of mass. What you need is the moment of inertia about the axis of rotation. (Use the parallel axis theorem.)

Sorry for not spotting that earlier!

 

[Becca93]

They give the moment of inertia about the center of mass. What you need is the moment of inertia about the axis of rotation. (Use the parallel axis theorem.)

Sorry for not spotting that earlier!

Alright, so,

Ip = Icm +mr^2
Ip = 0.039 + (2.3 x 0.57^2)
Ip = 0.78627

Plugging that into the E1 = E2 business earlier, I get w = 4.157 rad/s
Then v = wr,
v = 2.37 m/s
And again it is incorrect.

I honestly have no idea how to move forward.

 

[Doc Al]

Alright, so,

Ip = Icm +mr^2
Ip = 0.039 + (2.3 x 0.57^2)
Ip = 0.78627

Good.

Plugging that into the E1 = E2 business earlier, I get w = 4.157 rad/s

Redo this. I get a different value for ω.

 

[Becca93]

Good.


Redo this. I get a different value for ω.

I've finally got it! Thank you so very much!

 

[Doc Al]

Whew! You are most welcome.