How to find angular acceleration

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To find angular acceleration, the discussion highlights the importance of understanding the notation used, specifically the concept of epsilon (ε) representing an infinitesimally small time after t=0. It clarifies that since acceleration is constant over the 7 seconds, the value at t=0+ε can be treated as the same as at any other point in that interval. The calculations for tangential and centripetal acceleration are discussed, emphasizing the need for correct arithmetic and understanding vector directions. The confusion around the use of "atan" is resolved by clarifying that it should refer to the arctangent function, and a right triangle can help visualize the relationship between tangential and centripetal accelerations. Overall, the conversation focuses on ensuring accurate calculations and interpretations of angular motion concepts.
Sneakatone
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a)t=0 t=7
190/60*2pi=19.8 2950/60*2pi=308
308/7=41

I don't 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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Sneakatone said:
a)t=0 t=7
190/60*2pi=19.8 2950/60*2pi=308
308/7=41

I don't know what to do with acceleration with t=0.
You are told the acceleration is constant over the 7 seconds. The question asks for the acceleration at t=0+ε.
 
what do you mean? what does the E stand for?
 
Sneakatone said:
what do you mean? what does the E stand for?
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.
 
so can I calculate the acceleration at 1 sec?
 
Sneakatone said:
so can I calculate the acceleration at 1 sec?
What's to calculate? The acceleration is constant, and you already know what it is .
 
so acceleration is the same being 41.28?
 
That's it.
 
I have part a and b
for part c I used atan(atan/acen) to find angle but its incorrect.
 
  • #10
Sneakatone said:
for part c I used atan(atan/acen) to find angle but its incorrect.
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
for t=0 I did 7.4(7.4/71.2)=0.76
t=7 7.4(7.4/17177)=0.0031
 
  • #12
Sneakatone said:
for t=0 I did 7.4(7.4/71.2)=0.76
t=7 7.4(7.4/17177)=0.0031

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
so I am suppose tan-1(7.4/71.2) ?
 
Last edited:
  • #14
Sneakatone said:
so I am suppose tan-1(7.4/71.2) ?

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

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