A copper block rests 17.4 cm from the center of a steel turntable. The coefficient of static friction between block and surface is 0.465. The turntable starts from rest and rotates with a constant angular acceleration of 0.406 rad/s 2 . The acceleration of gravity is 9.8 m/s 2 . After what interval will the block start to slip on the turntable

12.61 s

Explanation:

Given that

Distance from the center if the turntable, r = 17.4 cm = 0.174 m

Coefficient of static friction, μ = 0.465

Angular acceleration, α = 0.406 rad/s²

Acceleration due to gravity, g = 9.8 m/s²

We know that

F_max = μmg

Also, we know that

F = mω²r

Now, for slip to occur, both forces must be equal to one another, and thus

mω²r = μmg

ω²r = μg

ω² = μg/r

ω² = (0.465 * 9.8)/0.174

ω² = 4.557 / 0.174

ω² = 26.19

ω = √26.19

ω = 5.12 rad/s

t = ω/α

t = 5.12/0.406

t = 12.61 s.

Thus, after 12 seconds, the block will start to slip on the turntable

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2