a penny is placed .1 m from the center of a turntable. if the coefficient of static friction between the penny and the turntable is .5, the maximum linear speed at which the penny can travel without slipping is approximately

a) .49m/s
b) .7 m/s
c) 1.3 m/s
d) 1.4 m/s
e) .2 km/s

explain please

thanks

The friction on the penny will equal centripetal force. At that equality, then linear speed is w*radius.

but.. what formula do i use?

ohh... nevermind!

To find the maximum linear speed at which the penny can travel without slipping, we need to use the concept of static friction. The maximum static friction is given by the equation:

f_s(max) = μ_s * N

where f_s(max) is the maximum static friction, μ_s is the coefficient of static friction, and N is the normal force.

In this case, the normal force acting on the penny is equal to its weight, as the penny is not accelerating vertically. The weight is given by:

N = m * g

where m is the mass of the penny and g is the acceleration due to gravity.

Now, to determine the maximum linear speed, we can equate the centripetal force required to keep the penny moving in a circle (mv^2/r) to the maximum static friction:

mv^2/r = μ_s * m * g

To cancel out the mass, we can then divide both sides of the equation by m:

v^2/r = μ_s * g

Finally, solving for v, we get:

v = sqrt(μ_s * g * r)

Given that the coefficient of static friction (μ_s) is 0.5, the acceleration due to gravity (g) is approximately 9.8 m/s^2, and the distance (r) from the center of the turntable to the penny is 0.1 m, we can substitute these values into the equation:

v = sqrt(0.5 * 9.8 * 0.1)
v = sqrt(4.9)
v ≈ 2.21 m/s

Therefore, the maximum linear speed at which the penny can travel without slipping is approximately 2.21 m/s. However, none of the provided answer choices match this value, so there might be a mistake or a rounding error in the problem statement or answer choices.