How Many Revolutions Before the Mass Slips on a Rotating Disk?

[hatcheezy]

Homework Statement

A 25gm mass sits on the surface of a circular disk 10 cm from the center of the disk. The force of static frictio nbetween the mass and surface of the disk is 0.075 N. The disk is initially at rest, but then experiences a constant angular acceleration of 0.80 rad/s^2. How many revolutions will the disk make before the mass starts to slip outward?

Homework Equations

i'm not exactly sure what equation to use-which is why I'm here...

The Attempt at a Solution

(.80rad) (1 rev) ...= 1.3rev/s
...5^2...2$$\pi$$rad

i don't know ehre to go from here...

 

[LowlyPion]

Consider the centripetal acceleration on the mass. At what ω then will the centripetal force on the mass overcome the friction that maintains it in position?

Once you know ω, you can then determine the angular displacement.

Maybe this link would be helpful:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

 

[nlightner]

I would approach this problem by first identifying the relevant equations and principles related to rotational motion and gravity. The main equation we can use in this scenario is Newton's second law for rotational motion, which states that the net torque on an object is equal to its moment of inertia multiplied by its angular acceleration. We can also use the equation for centripetal force, which relates the mass, speed, and radius of an object in circular motion.

In this problem, we are given the mass, distance from the center, and force of static friction, and we are asked to find the number of revolutions before the mass starts to slip. To do this, we can set up the equation for net torque and solve for the angular acceleration:

Στ = Iα

The torque in this case is provided by the force of static friction, which is equal to the product of the mass and the acceleration due to gravity (since the mass is not accelerating in the vertical direction). The moment of inertia for a disk is equal to 1/2 MR^2, where M is the mass of the disk and R is the radius. So we have:

0.075 N = (0.025 kg)(9.8 m/s^2)(10 cm/100 m)(1/2)(0.1 m)^2 α

Solving for α, we get:

α = 0.60 rad/s^2

Now, we can use the equation for angular acceleration to find the time it takes for the disk to reach this angular velocity:

ω = ω0 + αt

0.80 rad/s^2 = 0 + (0.60 rad/s^2)t

t = 1.33 seconds

Since we know the angular velocity and the initial angular position (0 radians), we can use the equation for angular displacement to find the number of revolutions:

θ = θ0 + ω0t + 1/2 αt^2

θ = 0 + 0 + 1/2(0.60 rad/s^2)(1.33 s)^2

θ = 1.00 rad

Since 1 revolution is equal to 2π radians, the number of revolutions is equal to 1.00 rad/(2π rad/rev) = 0.16 revolutions. Therefore, the disk will make approximately 0.16 revolutions before the mass starts to slip outward.