Physics Help: Tangential Acceleration. The speed of a point on a rotating turntable, which is 0.175 m from th?
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Physics Help: Tangential Acceleration. The speed of a point on a rotating turntable, which is 0.175 m from th?

[From: ] [author: ] [Date: 14-02-25] [Hit: ]
14 m/s² = 0.373 m/s² ----- (1) at t1 = 0.64 s V(t1) = 0 + a(tan)*(t1) = (0.373)*(0.64) = 0.239 m/s radius of the circle = r = 0.......
Here is the picture of the problem.
*There's two parts


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There is no picture.
What do you mean by "0.175 m from th" ?
Tangential acceleration :
Initial speed = Utan = 0
Final speed = Vtan = 0.798 m/s
time interval = t = 2.14 s
Tangential acceleration = a(tan) = (Vtan - Utan)/t
= (0.798 - 0)/2.14 m/s² = 0.798/2.14 m/s² = 0.373 m/s² ----- (1)
at t1 = 0.64 s
V(t1) = 0 + a(tan)*(t1) = (0.373)*(0.64) = 0.239 m/s
radius of the circle = r = 0.175 m
Centripetal acceleration = a(cent) = V'(t1)²/r
= (0.239)²/0.175 m/s² = 0.326 m/s² -------- (2)
Total acceleration = a(total) = √[{a(tan)}² + {a(cent)}²]
[a(tan) and a(cent) are at right angles to each other.]
Calculate
1
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