# Rotational motion

1. Jul 7, 2012

### Saitama

1. The problem statement, all variables and given/known data
http://i46.tinypic.com/548uox.png

2. Relevant equations

3. The attempt at a solution
My teacher said that we will take the axis at the point which is in contact with the ground (i don't get why we need to do so?). I followed the teacher's hint. Angular acceleration α=a/R, where a is the linear acceleration and R is the distance of the point from the axis.
Since α is same for all points,
$$\frac{a_1}{R_1}=\frac{a_2}{R_2}$$
a1=3m/s^2
a2=we need to find this.
R1=1m
R2=4/3m.
Using these values, i get:
$$a_2=4m/s^2$$
But this is wrong, the answer is 5m/s^2. I don't understand where i am going wrong.

Last edited by a moderator: Jul 7, 2012
2. Jul 7, 2012

### ehild

The problem asks the magnitude of the total acceleration of that point (centripetal included).

ehild

3. Jul 7, 2012

### Saitama

For the centripetal acceleration, i need the velocity of that point. As i did for linear acceleration, the linear velocity would be 4m/s.
Therefore, centripetal acceleration is $\frac{v^2}{R}=\frac{4*4*3}{4}=12m/s^2$
R=4/3 m
Net acceleration is $\sqrt{12^2+4^2}$ but this is not equal to 5. :(

4. Jul 7, 2012

### Yukoel

Hello Pranav-arora,
You are taking measurements from the instantaneous center of rotation right?First of all to confirm whether the given answer is correct or not by treating the same from the center of mass and ground frame.I think it should be correct.(though I haven't calculated)
As far as centripetal acceleration is concerned from the bottom most point the given point doesn't exactly cover a circular path on the plane w.r.t it.The curve should be a cycloid adjusted or say translated above the ground by a height of 2R/3 I suppose.So the formula for centripetal acceleration is $\frac{v^2}{r}$ where r denotes the radius of curvature of the curve at the point.So the calculation of centripetal acceleration done by you is incorrect unless you calculate the radius of curvature correctly.Correct me if I am wrong.
regards
Yukoel

Last edited: Jul 7, 2012
5. Jul 7, 2012

### ehild

Pranav, unlike your teacher I would put the axis of rotation through the centre of the wheel. It is the real axis, isn't it?

As Yukoel wrote, the points of the wheel do not move along circles around the instantaneous axis.
The radius will change, and the expression of acceleration in such system of reference is should include the derivatives

It is sure that you can describe the motion of any point of the wheel with respects to the axis through the centre (which is the CM of the wheel).
If you count the angle of rotation from the upper vertical, the components of the point at distance r=1/3 m from the centre of the wheel

x=XCM+rsin(φ)
y=R+rcos(φ)

The velocity components are:

vx=VCM+rωcos(φ)
vy=-rωsin(φ)

The rolling condition holds, so the magnitude of the linear velocity of the rim is v=Rω=VCM. As there is acceleration, the angular velocity changes with time.
According to the rolling condition, the angular acceleration is dω/dt=aCM/R.

Derivating the velocity components and substituting aCM/R for dω/dt you get the components of the acceleration.

ax=aCM+r(dω/dt)cos(φ)-rω2sin(φ)
ay=-r(dω/dt)sin(φ)-rω2cos(φ)

substituting for (dω/dt),

ax=aCM+raCM/Rcos(φ)-rω2sin(φ),

ay=-raCM/R sin(φ)-rω2cos(φ).

These are the general formulae. You need those for φ=0. What do you get for ax and ay?

ehild

6. Jul 7, 2012

### Saitama

Thanks you ehild and Yukoel for your replies!
How do you get these equation ehild? :uhh:
ax=aCM+raCM/R

ay=-rω2

If i use these equations, i get my answer to be 5m/s^2.

Yukoel said something about the radius of curvature, i don't understand how should i go about calculating it.

7. Jul 7, 2012

### Infinitum

I would solve this using a different method.

The acceleration you found in the first post can be used, and it will account for the tangential acceleration. Now the radial acceleration can be found by using the relative velocity between the axis of rotation and the given point. Then add them vectorially.

8. Jul 7, 2012

File size:
7.5 KB
Views:
51
9. Jul 7, 2012

### ehild

Pranav did exactly that, using the instantaneous axis, as his teacher suggested, for the axis of rotation.

ehild

10. Jul 7, 2012

### Infinitum

I took the center as the axis of rotation.

11. Jul 7, 2012

### ehild

Everything is all right and easy if you describe the motion of the point with respect to the centre (CM). It becomes complicated when somebody tries to find the acceleration in a frame of reference connected to the instantaneous axis. I never do it.

ehild

12. Jul 7, 2012

### Infinitum

Very true

13. Jul 7, 2012

### Saitama

Thank you ehild for that image, i get it now.

Any reason why i need to use the relative velocity here?

14. Jul 8, 2012

### Yukoel

Hello Pranav,
I think what Infinitum means is that you need to superimpose velocities/accelerations due to c.o.m and those relative to it.
I have drawn a diagram for a 2 dimensional case in the image below.
http://img842.imageshack.us/img842/3331/da0822968d2847a9bda0d4d.png [Broken]
Hoping this helps.
regards
Yukoel

Last edited by a moderator: May 6, 2017