# Is My Calculation of Centripetal Acceleration and Tangential Velocity Correct?

• user079622
user079622
Homework Statement
I must find centripetal acceleration and tangential velocity, I post this to check my results.
Relevant Equations
vt = ω r
ac = ω2 r
Right picture is two turn tables on of top of the other, smaller turn table is connected with shaft to bigger one so it rotate around itself and in same time "revolve" around center of bigger one which is also rotate about itself. They both rotate clockwise.

I observe case from inertial frame of reference, earth. I must find centripetal acceleration and tangential velocity in point B.

r1=1m ω1=1000RPM
r2=10 ω2=500RPM

Tangential velocity in point B = tangen. velocity in point P1(right picture) + tangen. velocity in point A
vt = ω r
A= 104.72 m/s
P1=523.6 m/s

They are two parallel velocity vectors pointing in same direction so I add them to get tangential velocity in point B=628.32 m/s

Centripetal acceleration in point B = cent. accleration in point P1(right picture) + cent. acceleration in point A

ac = ω2 r
P1= 27416 m/s2
A= 10966 m/s2

They are two vectors that point in same direction,(toward left at my graph), I add them to get centripetal acceleration in point B= 38382 m/s2

Is my logic and results correct?
Point A show case where smaller turn table rotate just around itself, revolution is zero)

Last edited:
Looks fine to me.

haruspex said:
Looks fine to me.
Do you maybe have any doubts about the way how I calculated centripetal acceleration?

user079622 said:
Do you maybe have any doubts about the way how I calculated centripetal acceleration?
Let me think on it some more…
A sanity check: consider ##\omega_1=\omega_2##.
Now B is always at ##r_1+r_2## from ##P_2##, so its acceleration is clearly ##(r_1+r_2)\omega^2##. Your method gives ##r_1\omega^2+r_2\omega^2##, which is the same.
Try ##\omega_1=0##.

haruspex said:
Let me think on it some more…
A sanity check: consider ##\omega_1=\omega_2##.
In this case is easy and clear.

Last edited:
So are you reassured?

haruspex said:
So are you reassured?
If ω1=ω2 then vt=576 m/s and acc=30159 m/s2
This is moon case, like small table is locked so it cant rotate in relation to big one.

Last edited:
user079622 said:
If ω1=ω2 then vt=576 m/s and acc=30159 m/s2
This is moon case, like small table is locked so it cant rotate in relation to big one.
Yes, but the point of considering such special cases is to see whether your approach in post #1 works. Don't plug in any numbers. I showed in post #4 that your method gives the right formula for that case. Can you do the same for ##\omega_1=0##?

haruspex said:
Yes, but the point of considering such special cases is to see whether your approach in post #1 works. Don't plug in any numbers. I showed in post #4 that your method gives the right formula for that case. Can you do the same for ##\omega_1=0##?
Formula for my case is r1 (ω1)2 + r2 ( ω2)2

I must think...
If ω1=0, that mean small table not rotate from earth frame,so point B always point toward right side
If is not rotate then acc=0 from rotation, so it has only acceleration from "revolution" around point P2, that is equal
(r1+r2 ) (ω2)2

But small table then rotate in relation to big table at -500RPM.
Am I correct?

user079622 said:
If is not rotate then acc=0 from rotation, so it has only acceleration from "revolution" around point P2, that is equal
(r1+r2 ) (ω2)2
That would conflict with your general solution. Plugging ##\omega_1=0## into ##r_1\omega_1^2+r_2\omega_2^2## gives ##r_2\omega_2^2##, not ##(r_1+r_2)\omega_2^2##.
But fortunately your reasoning in post #9 is wrong. If ##\omega_1=0##, what point is B orbiting?

haruspex said:
That would conflict with your general solution. Plugging ##\omega_1=0## into ##r_1\omega_1^2+r_2\omega_2^2## gives ##r_2\omega_2^2##, not ##(r_1+r_2)\omega_2^2##.
But fortunately your reasoning in post #9 is wrong. If ##\omega_1=0##, what point is B orbiting?
Yes I miss orbit path
B orbit around point P2+r1
so acc= r2 (ω2)2

Last edited:
user079622 said:
Yes I miss orbit path
B orbit around point P2+r1
so acc= r2 (ω2)2
Right. So your solution in post #1 looks to be good.

haruspex said:
Right. So your solution in post #1 looks to be good.
Thanks for help.

## What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is responsible for changing the direction of the object's velocity without necessarily changing its speed. The formula for centripetal acceleration is $$a_c = \frac{v^2}{r}$$, where $$v$$ is the tangential velocity and $$r$$ is the radius of the circular path.

## How is tangential velocity different from centripetal acceleration?

Tangential velocity refers to the linear speed of an object moving along the circular path, measured at any point tangent to the circle. It is given by $$v = r \omega$$, where $$r$$ is the radius and $$\omega$$ is the angular velocity. In contrast, centripetal acceleration is the rate of change of the direction of this tangential velocity, directed towards the center of the circle.

## How do you calculate the centripetal force acting on an object?

The centripetal force can be calculated using the formula $$F_c = m \cdot a_c$$, where $$m$$ is the mass of the object and $$a_c$$ is the centripetal acceleration. Substituting the formula for centripetal acceleration, it becomes $$F_c = m \cdot \frac{v^2}{r}$$. This force is necessary to keep the object moving in a circular path.

## What happens to the centripetal acceleration if the radius of the circular path is doubled?

If the radius of the circular path is doubled, the centripetal acceleration will be halved, assuming the tangential velocity remains constant. This is because centripetal acceleration is inversely proportional to the radius, as given by the formula $$a_c = \frac{v^2}{r}$$.

## Can an object have both centripetal acceleration and tangential acceleration simultaneously?

Yes, an object can have both centripetal and tangential accelerations simultaneously. Centripetal acceleration is directed towards the center of the circular path, changing the direction of the velocity. Tangential acceleration, on the other hand, is along the tangent to the path and changes the magnitude of the tangential velocity. This occurs, for example, in non-uniform circular motion where the speed of the object is changing.

Replies
17
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
1
Views
3K
Replies
2
Views
2K
Replies
8
Views
1K
Replies
35
Views
3K
Replies
15
Views
2K