Auburn2017
- 59
- 1
This is just not making sense to me.TSny said:Speed doesn't have a direction.
This is just not making sense to me.TSny said:Speed doesn't have a direction.
but you can't get acceleration from speedTSny said:Speed is the magnitude of the velocity. Velocity is a vector. Speed is a scalar.
So, the velocity of C is not the same as the velocity of A. But the speed of C does equal the speed of A.
how would you find the angular acceleration and velocity of the wheel?TSny said:You can get the magnitude of the normal acceleration from the speed and the radius. Normal acceleration is also called centripetal acceleration.
rA=rB+rA/B
vA=vB+vA/B
aA=aB+aA/B
vA/B=ω×rA/B
aA/B=(aA/B)n+(aA/B)t
(aA/B)n=ω×(ω×rA/B)
(aA/B)t=α×rA/B
Yes I am familiar with such equationTSny said:Are you familiar with the formula acentripetal = v2/r ?
Or are you required to solve this problem using only the equations that you listed:
49/6TSny said:Great! What do you get for the magnitude of the centripetal acceleration?
brain fart just 49/6TSny said:Why the square root?
the magnitude of the normal acceleration would be radius times angular acceleration which we know from point B correct?TSny said:OK. All you need to do is put it all together now.
But what is the value of anTSny said:Yes, but why go there? You already have values for an and at. Use these in your earlier expressions to obtain the normal and tangential acceleration vectors in terms of i and j. For example, see your post #21.
SAMETSny said:How does the magnitude of the tangential acceleration of C compare to the magnitude of the tangential acceleration of B?
When I plug and chug I do not get the answer :/TSny said:Yes.