How to find angular acceleration

1. Apr 11, 2013

Sneakatone

a)t=0 t=7
190/60*2pi=19.8 2950/60*2pi=308
308/7=41

I dont know what to do with acceleration with t=0.

b)for tangental I used the equation alpha*R

for centripetal I used w^2*R

c) I think the equation might be applied
atan(atan/acen)

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2. Apr 11, 2013

haruspex

You are told the acceleration is constant over the 7 seconds. The question asks for the acceleration at t=0+ε.

3. Apr 11, 2013

Sneakatone

what do you mean? what does the E stand for?

4. Apr 11, 2013

haruspex

It's epsilon, a fairly standard notation for an arbitrarily small (usually positive) quantity. In other words, it isn't asking for the acceleration exactly at t=0 (which would be impossible to determine) but the acceleration just a fraction later.
You are told the acceleration is constant over the 7 seconds, but that could mean over the interval [0, 7] or [0, 7) or (0, 7] or (0, 7). The point is that 0+ε lies in all of them, whereas 0 only lies in the first two.

5. Apr 11, 2013

Sneakatone

so can I calculate the acceleration at 1 sec?

6. Apr 11, 2013

haruspex

What's to calculate? The acceleration is constant, and you already know what it is .

7. Apr 11, 2013

Sneakatone

so acceleration is the same being 41.28?

8. Apr 11, 2013

haruspex

That's it.

9. Apr 11, 2013

Sneakatone

I have part a and b
for part c I used atan(atan/acen) to find angle but its incorrect.

10. Apr 11, 2013

haruspex

That's the right principle. Maybe you made an arithmetic error, or perhaps you need to consider that the radius, as a vector, points outwards, whereas the acceleration points somewhat inwards. (I.e., the centripetal acceleration is negative).

11. Apr 11, 2013

Sneakatone

for t=0 I did 7.4(7.4/71.2)=0.76
t=7 7.4(7.4/17177)=0.0031

12. Apr 11, 2013

haruspex

You seem to have confused yourself by an overloaded abbreviation. You wrote previously "atan(atan/acen)", which I interpreted as arctan(atan/acen), since that is correct, but in your numerical working you have interpreted it as atan(atan/acen)

13. Apr 12, 2013

Sneakatone

so im suppose tan-1(7.4/71.2) ?

Last edited: Apr 12, 2013
14. Apr 12, 2013

haruspex

Yes. Do you see why? Draw a right-angled triangle with the side opposite angle theta representing the tangential acceleration, 7.4, and the (non-hypotenuse) side adjacent representing the centripetal acceleration, 71.2. What would be the value of theta?

15. Apr 12, 2013

Sneakatone

yes i do ,aslo my calculator was in radians at first so I got confused, Thanks for the help!

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