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Rotational Kinematics

Free Homework Help!PhysicsRotational Kinematics

Steven Dang
asked 9 months ago

A bug is on the rim of a disk of diameter 15 in that moves from rest to an angular speed of 71 rev/min in 5.2 s.

1. What is the tangential acceleration?

2.What is the tangential velocity of the bug at the end of the 5.2 s? 

3. One second after the bug starts from rest, what is its total acceleration? 

4. What angle does this acceleration make with ac? 

1 Answers
Alexander Chioma Staff answered 9 months ago

Hello! Thank you for the question.
The first thing we need to find for this problem is the angular velocity of the bug in rads per second.
w = (71rev/min) (1min/60s) (2(pi)rad/1rev)
w = 7.43rad/s
a)     The tangential acceleration a is given by a=ra. We want this radius in meters.
r = (7.5in) (2.54cm/1in) = 19.05cm = 0.1905m
ara = rw/t      =     (0.1905m) (7.43rad/s) / (5.2s) = 0.272 m/s^2
b)      The tangential velocity v is given by v=rw
v = (0.1905m) (7.43rad/s) = 1.416m/s
c)     First we need the angular velocity w after 1 second
w = wi + at = 0 + (7.43rad/s / 5.2s) (1s) = 1.429rad/s
Radial acceleration ac = rw^2 = (0.1905m) (1.429rad/s)^2 = 0.389rad/s^2
Now total acceleration Atot is given by the distance formula with ac and a from earlier.
Atot = sqrt[(0.272)^2 + (0.389)^2] = 0.475m/s^2
d)      The angle x that the total acceleration makes with ac is given by
x = tan^-1(a/ac) = tan^-1(0.272/0.389) = 34.96degrees

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