Blocks on turntable with friction = hard problem

In summary, two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 4 cm from the center and the outer block is 5 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.71, and the string is taut. The maximum angular frequency for neither block to slide is 12.4 rad/s. For part B, with the blocks having a mass of 34 g, the tension in the string for this maximum angular frequency is unknown and may require further calculations.
  • #1
yopy
43
0
Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 4 cm from the center and the outer block is 5 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.71, and the string is taut.

a) What is the maximum angular frequency such that neither block slides?

b) Now suppose that the blocks each have a mass m = 34 g. For the value of w you just found, what is the tension in the string?


i worked on part A for a long time and found the answer to be 12.4 rad / s

i am having problems calculating part B, can anyone point me in the right direction, thanks
 
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  • #2
I get a different answer for part a).

Perhaps you can show your work?
 
  • #3


For part B, you will need to use the equation for centripetal force, Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the block, v is the velocity, and r is the radius.

First, you will need to find the velocity of the blocks using the angular frequency you found in part A. The velocity can be calculated using the formula v = ωr, where ω is the angular frequency and r is the radius.

Once you have the velocity, you can plug it into the centripetal force equation along with the mass of the blocks (m = 34 g) and the radius (r = 4 cm). This will give you the centripetal force acting on the blocks.

Since the blocks are not sliding, the centripetal force must be equal to the tension in the string. Therefore, the tension in the string can be calculated by setting the centripetal force equal to the tension.

T = mv^2/r = (0.034 kg)(v^2)/(0.04 m)

You already know the value of v from part A, so you can plug that in and solve for the tension (T).

T = (0.034 kg)(12.4 rad/s)^2/(0.04 m) = 13.6 N

Therefore, the tension in the string is 13.6 N when the maximum angular frequency is 12.4 rad/s.
 

1. What is the "Blocks on turntable with friction = hard problem"?

The "Blocks on turntable with friction = hard problem" is a well-known physics problem that involves a turntable with two blocks placed on top of it. The turntable is rotating at a constant angular velocity, and there is friction between the blocks and the turntable. The problem is to determine the motion of the blocks and the turntable over time.

2. Why is this problem considered to be difficult?

This problem is considered difficult because it involves multiple variables and forces that interact with each other. The friction between the blocks and the turntable adds a complex component to the problem, making it challenging to solve analytically.

3. What are some real-world applications of this problem?

This problem has many real-world applications, such as in the design of rotating machinery and vehicles. For example, engineers may use similar principles to analyze the stability of a rotating crane or the motion of a car on a curved road.

4. What are some strategies for solving this problem?

One strategy for solving this problem is to break it down into smaller, simpler problems. For example, you can analyze the motion of each block separately and then combine the results. Another strategy is to use numerical methods, such as computer simulations, to approximate the solution.

5. Can this problem be solved exactly?

It is possible to solve this problem exactly, but it requires advanced mathematical techniques and assumptions. In most cases, an exact solution is not necessary, and approximations or numerical methods can provide accurate results.

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