Asked by joy
Harry Potter decides to take Pottery 101 as an elective to satisfy his arts requirement at Hogwarts. He sets some clay
(m = 4.00 kg)
on the edge of a pottery wheel
(r = 0.730 m),
which is initially motionless. He then begins to rotate the wheel with a uniform acceleration, reaching a final angular speed of 2.800 rev/s in 2.80 s.
(a) What is the speed of the clay when the initial 2.80 s has passed?
m/s
(b) What is the centripetal acceleration of the clay initially and when the initial 2.80 s has passed? (Enter the magnitudes of your answers.)
ac,i = m/s^2
ac,f = m/s^2
(c) What is the magnitude of the constant tangential acceleration responsible for starting the clay in circular motion?
m/s^2
(m = 4.00 kg)
on the edge of a pottery wheel
(r = 0.730 m),
which is initially motionless. He then begins to rotate the wheel with a uniform acceleration, reaching a final angular speed of 2.800 rev/s in 2.80 s.
(a) What is the speed of the clay when the initial 2.80 s has passed?
m/s
(b) What is the centripetal acceleration of the clay initially and when the initial 2.80 s has passed? (Enter the magnitudes of your answers.)
ac,i = m/s^2
ac,f = m/s^2
(c) What is the magnitude of the constant tangential acceleration responsible for starting the clay in circular motion?
m/s^2
Answers
Answered by
bobpursley
I am reluctant to do these without seeing your work.
a. SPEED=w*r=2PI*2.800*radius
b. initially? isn't it zero? the wheel is at stop.
final? w^2*r=(2PI*2.8)^2* radius
c. tangential acceleration=changeradialvelociy/time
= 2PI*2.8*radius/time
a. SPEED=w*r=2PI*2.800*radius
b. initially? isn't it zero? the wheel is at stop.
final? w^2*r=(2PI*2.8)^2* radius
c. tangential acceleration=changeradialvelociy/time
= 2PI*2.8*radius/time
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