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
2 answers:
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
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.
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