Question

A copper block rests 38.9 cm from the center of a steel turntable. The coefficient of static friction between the block and the surface is 0.40. The turntable starts from rest and rotates with a constant angular acceleration of 0.60 rad/s2 . After what time interval will the block start to slip on the turntable

Answer 1

Answer:

t = 5.291 s

Explanation:

The copper block starts moving when the force responsible for the circular motion matches the frictional force.

The frictional force is given by μN = μ mg where N = Normal reaction on the block.

μ = coefficient of static friction = 0.4

The force causing circular motion is given by mv²/r = mrw²

mrw² = 0.4 mg

rw² = 0 4g

r = 38.9 cm = 0.389 m

g = 9.8 m/s²

w = √(0.4×9.8/0.389)

w = 3.174 rad/s

We then use equation of motion for the turntable to find out when its angular velocity reaches 3.174 rad/s which is enough to move the block

w = w₀ + αt

w = 3.174 rad/s

w₀ = 0 rad/s ( the turntable was initially at rest)

α = 0.60 rad/s²

t = ?

3.174 = 0 + 0.6t

t = 5.291 s

Answer 2

Explanation:

Fmax = μFn

Where,

Fn = normal force

= mg

μ = coefficient of static friction

Fmax = 0.4mg

Angular force, Fi = mω^2r

mω^2 * r = 0.4mg

ω^2 * r = 0.4g

ω = sqrt(0.4 * 9.81/0.389)

= 3.18 rad/s

Using equations of angular motion,

ωf = ωi + αt

ωi = 0 rads

t = ωf/α

= 3.18/0.6

= 5.29 s.