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

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.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft