Kinematics of Rigid Body Plane Motion

[Auburn2017]

Homework Statement

The wheel shown rotates about point O.
Point A has a velocity=-7j in/sec.
Point B has a tangential velocity=-4i in/sec^2
Determine the absolute acceleration of Point C located on the circle.
radius=6 in
You will have to look at the figure for more clarification please.

Homework Equations

r_A=r_B+r_A/B
v_A=v_B+v_A/B
a_A=a_B+a_A/B
v_A/B=ω×r_A/B
a_A/B=(a_A/B)_n+(a_A/B)_t
(a_A/B)_n=ω×(ω×r_A/B)
(a_A/B)_t=α×r_A/B

The Attempt at a Solution

I tried finding the velocity of C relative to A. I feel like there is an easy way to find the angular acceleration and angular velocity of the wheel but I don't really know how. Thanks for your help.

 

[TSny]

The wheel is considered to be a rigid body rotating about a fixed axis. How does the absolute speed of point C compare to the absolute speed of A? How does the magnitude of the absolute tangential acceleration of C compare to that of B?

 

[Auburn2017]

The wheel is considered to be a rigid body rotating about a fixed axis. How does the absolute speed of point C compare to the absolute speed of A? How does the magnitude of the absolute tangential acceleration of C compare to that of B?

Since all three points are on the edge of the wheel then they should all have the same absolute acceleration and velocity.

 

[TSny]

The magnitudes of the velocity and tangential acceleration will be the same. Directions?

 

[Auburn2017]

The magnitudes of the velocity and tangential acceleration will be the same. Directions?

velocity CW
acceleration CCW

 

[TSny]

OK. But more specifically, can you express the tangential acceleration vector of point C in terms of the unit vectors $\hat{i}$ and $\hat{j}$?

 

[Auburn2017]

No I can't. Isn't that what the problem is asking?

 

[TSny]

No, there's more to the problem than just doing that. Does point C have any other component of acceleration besides tangential?

 

[Auburn2017]

No, there's more to the problem than just doing that. Does point C have any other component of acceleration besides tangential?

Yes it has a normal acceleration toward the center of the circle. I guess the tangential acceleration would be a_tcos60° i and a_tsin60° j

 

[TSny]

You need to draw the tangential and normal acceleration vectors for point C on your diagram. The tangential acceleration of C is not vertically upward (i.e., not +j) and it's not "to the right" (i.e., not i).

Don't worry about the "hats". You could use bold type instead.

 

[Auburn2017]

You need to draw the tangential and normal acceleration vectors for point C on your diagram. The tangential acceleration of C is not vertically upward (i.e., not +j).

Don't worry about the "hats". You could use bold type instead.

look at my edited answer please

 

[TSny]

Yes it has a normal acceleration toward the center of the circle. I guess the tangential acceleration would be a_tcos60° i and a_tsin60° j

That's getting close. But you need to check if 60^o is correct here.

 

[Auburn2017]

That's getting close. But you need to check if 60^o is correct here.

yes it is correct

 

[TSny]

Did you actually draw the tangential acceleration vector at C on your diagram? What angle does it make to the horizontal?

 

[Auburn2017]

Did you actually draw the tangential acceleration vector at C on your diagram? What angle does it make to the horizontal?

Yes I drew it. I'm not sure how to tell the angle though...

 

[TSny]

You'll need to use a little geometry.

 

[Auburn2017]

You'll need to use a little geometry.

I'm going with 30 degrees

 

[TSny]

Did you work it out, or is that a guess?

Look at the figure below. What is the angle $\phi$? Form $\phi$ do you see how to get $\theta$?

 

[Auburn2017]

Did you work it out, or is that a guess?

Look at the figure below. What is the angle $\phi$? Form $\phi$ do you see how to get $\theta$?
View attachment 103452

Yes theta would equal 30 degrees

 

[TSny]

Good. Repeat for the normal acceleration.

 

[Auburn2017]

normal acceleration would be a_ncos60°-i and a_nsin60°+j

 

[TSny]

Good.

 

[Auburn2017]

Good.

Am I correct in saying that I will need to know the velocity of point C to find the final answer?

 

[TSny]

The speed at C should be sufficient. Don't worry about the velocity (vector) at C.

 

[Auburn2017]

The speed at C should be sufficient. Don't worry about the velocity (vector) at C.

And I will get the speed by using the relative velocities of the points B and C

 

[TSny]

How does the speed at C compare to the speed at A?

 

[Auburn2017]

How does the speed at C compare to the speed at A?

it is the speed of A plus speed of C relative to A

 

[TSny]

You need the absolute speed of C. How does the absolute speed of C compare to the absolute speed of A? See posts #3 and #4.

 

[Auburn2017]

You need the absolute speed of C. How does the absolute speed of C compare to the absolute speed of A? See posts #3 and #4.

it is the same as A but in a different direction

 

[TSny]

Speed doesn't have a direction.

 

[Auburn2017]

Speed doesn't have a direction.

This is just not making sense to me.

 

[TSny]

Speed is the magnitude of the velocity. Velocity is a vector. Speed is a scalar.

So, the velocity of C is not the same as the velocity of A. But the speed of C does equal the speed of A.

 

[Auburn2017]

Speed is the magnitude of the velocity. Velocity is a vector. Speed is a scalar.

So, the velocity of C is not the same as the velocity of A. But the speed of C does equal the speed of A.

but you can't get acceleration from speed

 

[TSny]

You can get the magnitude of the normal acceleration from the speed and the radius. Normal acceleration is also called centripetal acceleration.

 

[Auburn2017]

You can get the magnitude of the normal acceleration from the speed and the radius. Normal acceleration is also called centripetal acceleration.

how would you find the angular acceleration and velocity of the wheel?