Question

Question :  A pulley one metre in diameter rotating at 600 r.p.m. is brought to rest in 80 sec. by a constant force of friction on its shaft. How many revolutions does it makes before coming to rest?

Answer 1

I think may be 7 is the answer

Answer 2

Begin the question by writing down all the known values and converting them to their standard form.

Given:

ω₁ (initial) = 600 rpm = $\frac{600 X 2π}{60}$

(∵ revolutions per minute can be converted to radians per second)

ω₂ (final) = 0

(∵ "brought to rest")

t = 80 seconds

To Find -

Θ ( from which we can easily calculate the number of revolutions)

Solution:

we know

ω₂-ω₁=αt

∴α=$\frac{20\pi }{80}$

α=$\frac{\pi }{4}$

next

ω₂²-ω₁²=2αΘ

∴Ф=$\frac{(20π)²}{2\alpha }$

Ф=800π (∵α= π/4)

∴ no. of revolutions are 800π÷2π

=400 revolutions