Pivoting Sledgehammer

  • Thread starter Becca93
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In summary, the sledgehammer has a mass of 2.30 kg, is connected to a frictionless pivot at the tip of its handle, has a moment of inertia about the center of mass of 0.0390 kgm2, and is released from rest at an angle of 40.0 degrees. When it passes through horizontal, the center of mass is going to have a speed of 36.1 m/s.
  • #1
Becca93
84
1
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.
 
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  • #2
Becca93 said:
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.)
 
  • #3
Doc Al said:
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?
 
  • #4
Becca93 said:
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.
 
  • #5
Doc Al said:
Check your arithmetic.

Oh, okay.
But I end up with 11.7, which is still the incorrect answer.
 
  • #6
Becca93 said:
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! :redface:
 
  • #7
Doc Al said:
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! :redface:

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.
 
  • #8
Becca93 said:
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.
 
  • #9
Doc Al said:
Good.


Redo this. I get a different value for ω.

I've finally got it! Thank you so very much!
 
  • #10
Whew! You are most welcome.
 

What is a pivoting sledgehammer?

A pivoting sledgehammer is a tool used for demolition and construction work. It is a heavy, long-handled hammer with a large metal head on one end and a flat striking surface on the other. The pivoting feature allows for more precise and controlled swings.

How is a pivoting sledgehammer different from a regular sledgehammer?

The main difference between a pivoting sledgehammer and a regular sledgehammer is the pivoting feature. This allows for more accuracy and control in the swing, making it easier to hit a specific target or angle. Regular sledgehammers have a fixed head and are typically used for more general demolition work.

What are the benefits of using a pivoting sledgehammer?

The pivoting feature of a sledgehammer allows for more precision and control in each swing. This can be beneficial for tasks such as breaking up concrete or driving in stakes. It can also help reduce strain on the user's arms and shoulders by allowing for a more natural swinging motion.

What materials can a pivoting sledgehammer be used on?

A pivoting sledgehammer can be used on various materials such as concrete, brick, wood, and metal. It is a versatile tool that can be used for demolition, construction, and landscaping tasks.

Are there different sizes of pivoting sledgehammers available?

Yes, there are different sizes of pivoting sledgehammers available, ranging from 6 pounds to over 20 pounds. The size of the sledgehammer needed will depend on the task at hand and the strength of the user. It is important to choose the right size for the job to avoid injury or strain.

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