( P.19) Tangential Acceleration

Click For Summary
The discussion focuses on calculating the acceleration of a point on the edge of a circular disk rotating at a constant angular speed of 5.0 revolutions per second. It clarifies that while the tangential acceleration is zero due to constant angular speed, the problem actually asks for centripetal acceleration. The centripetal acceleration is calculated using the formula a_c = rω^2, where ω must be converted to radians for accurate results. The conversion from revolutions to radians is necessary for the calculations to hold true. Ultimately, the centripetal acceleration at the edge of the disk is determined to be 9.9 m/s².
gcombina
Messages
157
Reaction score
3
A circular disk of radius 0.010 m rotates with a constant angular speed of 5.0 rev/s. What is the acceleration of a point on the edge of the disk?
answer is 9.9 m/s ^2

My attempt:
The problem is asking me for the tangential acceleration of the disk.
At = rα

so to solve the above equation i need the value of α

the problem just gave me the velocity , so can i use that velocity in lie of acceleration? why not!?
 
Physics news on Phys.org
gcombina said:
A circular disk of radius 0.010 m rotates with a constant angular speed of 5.0 rev/s. What is the acceleration of a point on the edge of the disk?
answer is 9.9 m/s ^2

My attempt:
The problem is asking me for the tangential acceleration of the disk.
At = rα

If the angular speed is constant, then α (angular acceleration) is zero, right?

(The derivative of a constant is zero.)
(In other words, angular speed is not changing.)

Therefore the tangential acceleration is zero.

gcombina said:
The problem is asking me for the tangential acceleration of the disk. 
The question is not asking for the tangential acceleration, it just asked for the acceleration.What other acceleration could it be talking about?EDIT:
It could be asking for tangential acceleration, but that's not what it's specifically asking for, it's just asking for whatever acceleration there is (whether it's in the tangent direction or some other direction)
 
Last edited:
it is asking for LINEAR acceleration right?
or CENTRIPETAL acceleration
I would totally say that because the disk is in constant speed and the acceleration of the whole disk is 0 m/s^2 but the answer is 9.9 m/s^2
 
Last edited:
gcombina said:
it is asking for LINEAR acceleration right?

It's asking for centripetal acceleration :)
 
aha,OK let me see now...
 
so I got for centripetal acceleration = rw^2
I converted 5.0 rev to rads so I got 31.414=5 rad/s

then I plug in

= (0.010 m) (31.415 /s)^2
= 9.9 m/s^2

Question
When I do the convertion from 5.0 to radians. Do I need to convert that 5.0 rev to radians??

for example i had
= (5.0 rev/s) (2 pi radians/1 rev)
= (5.0 rev/s) [(2) (3.1415 rad/1 rev)]
= (5.0 rev/s) (6.283 rad/1 rev)
= 31.415 rad/s (final answer) (do I really need to put say 1 rev = 2 pi rad?
 
gcombina said:
Question
When I do the convertion from 5.0 to radians. Do I need to convert that 5.0 rev to radians??

Yes. The centripetal acceleration formula is:

a_c=\frac{v^2}{r}=\frac{(ωr)^2}{r}=rω^2

If ω was measured in anything other than radians (like revolutions) then ωr=v would not be true.
(and so the above formula would not simplify to a_c=rω^2)
 

Similar threads

Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
26
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
8K