Finding Mass: Calculating Friction & Motion

[chenny]

A copper block rests 30.cm from the center of a steel turntable. The coefficient of static friction between the block and the surface is .53. The turntable starts from rest and rotates with a constant angular acceleration of .50 rad/s^2. After what time interval will the block start to slip on the turntable.
Hint: The normal force in this case equals the weight of the block.



Fc=u(mew)timesN



Tangential Speed = .15m/s
mew =.53
N = mg
N = m(9.81 m/s^2)

How do I find the mass?


Please and Thank you...

 

[NTUENG]

Hi chenny,

This is a rotational motion problem. By Newton's 2nd law, Net force= ma. F- Ff = ma. Since tangential acceleration = radius x angular acceleration, F- Ff = m (0.3m x 0.5 rad/s²). When block starts to slip, Ff = (meu)N = 0.53mg. Manipulate Newton's 2nd law, u get F = ma = m dv/dt and solve for t. But u cannot find the mass of the block, thus u cannot find the time when block starts to slip on turntable.

How to find the mass of the block?? Can someone pls help us??

 

[m2003]

To find the mass of the copper block, we can use the equation N = mg where N is the normal force, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2). Since the normal force is equal to the weight of the block, we can rearrange the equation to solve for the mass:

m = N/g

Substituting in the known values, we get:

m = (0.53)(9.81 m/s^2) / (9.81 m/s^2)

m = 0.53 kg

So the mass of the copper block is 0.53 kg.

Now, to find the time interval after which the block will start to slip on the turntable, we can use the equation for frictional force:

Ff = uN

Where Ff is the frictional force, u is the coefficient of static friction, and N is the normal force.

Since we know the coefficient of static friction and the normal force (which we found to be the weight of the block), we can calculate the maximum frictional force that the block can withstand before slipping:

Ff = (0.53)(0.53 kg)(9.81 m/s^2)

Ff = 2.74 N

Next, we can use the equation for centripetal force to find the maximum force that the turntable can exert on the block without causing it to slip:

Fc = mv^2/r

Where Fc is the centripetal force, m is the mass of the block, v is the tangential speed (which we were given as 0.15 m/s), and r is the distance from the center of the turntable (which was given as 30 cm or 0.3 m).

Substituting in the known values, we get:

2.74 N = (0.53 kg)(0.15 m/s)^2 / (0.3 m)

Solving for the mass, we get:

m = 0.53 kg

Therefore, we can conclude that at the instant when the centripetal force reaches 2.74 N, the block will start to slip on the turntable.

To find the time interval, we can use the equation for centripetal acceleration:

ac = v^2/r

Where ac is the centripetal acceleration, v is the tangential speed