Coefficient of dynamic friction

[suspenc3]

Homework Statement

A 4 kg collar C rides along a horizontally rotating arm AB. A spring k connects C to A. The arm AB rotates at a constant rate of 15 rpm. At the illustrated instant r = 1 m, r˙ = 1.1 m/s, and r¨ = 4 m/s2. The coefficient of dynamic friction, μk, is equal to 0.3. What is the magnitude of the total force exerted by the spring on the collar?

http://img337.imageshack.us/img337/958/dynamicsqig9.th.gif

Homework Equations

$$a=(\ddot {r} -r \dot {\theta}^2)i_R + (2 \dot {r} \dot {\theta} + r \ddot {\theta})i_ {\theta}$$

The Attempt at a Solution

$$\\sumF_r=ma_r=m(\ddot {r} -r \dot {\theta}^2)=-F_s-\mu R$$
$$\\sumF_{\theta}=ma_{\theta}=m(2 \dot {r} \dot {\theta} + r \ddot {\theta})=R$$

Everything is known exepth for $$F_s$$ which is the answer (force on spring), R, and $$\dot {\theta}$$.

$$\dot {\theta}=\frac {rpm*2 \pi}{60} = 1.57$$

I plug in numbers and get R=13.82 and Fs=-10.27

Any help would be appreciated

 

[anathema]

!

Thank you for your question. After reviewing your work, I noticed a few errors in your calculations. Firstly, the value for \dot {\theta} should be 1.57 rad/s, not 1.57 rpm. This is because the units for angular velocity are in radians per second, not revolutions per minute. Additionally, the value for R should be 4.82 N, not 13.82 N. This is because R is equal to the mass of the collar (4 kg) multiplied by the centripetal acceleration (1.2 m/s^2).

Using these corrections, we can calculate the force exerted by the spring on the collar. The sum of forces in the radial direction is given by:

\sum F_r = ma_r = m(\ddot {r} - r \dot {\theta}^2) = -F_s - \mu R

Plugging in the values for mass, acceleration, and R, we get:

4(-4) = -F_s - (0.3)(4.82)

Solving for F_s, we get:

F_s = 11.88 N

Therefore, the magnitude of the total force exerted by the spring on the collar is 11.88 N.

I hope this helps clarify your calculations. Let me know if you have any further questions. Keep up the great work in your studies!

 

[ogb p]

.

I would approach this problem by first reviewing the given information and equations to make sure I understand the scenario and the variables involved. From the problem statement, it is clear that the arm AB is rotating at a constant rate of 15 rpm and the collar C is connected to A by a spring with a known spring constant k. The coefficient of dynamic friction, μk, is also given and we are asked to find the magnitude of the force exerted by the spring on the collar.

Based on the equations provided, it seems that we are dealing with circular motion and therefore the equations for centripetal acceleration and force may be applicable. However, we also have to take into account the force of friction and the acceleration of the collar along the arm AB.

To solve for the force exerted by the spring on the collar, we can start by using the equation for centripetal acceleration:

a_r = \ddot {r} -r \dot {\theta}^2

Substituting the known values, we get:

a_r = 4 m/s2 - (1 m)(1.57 rad/s)^2 = 2.47 m/s2

Next, we can use the equation for the net force in the radial direction:

ΣF_r = ma_r = -F_s - μkR

Rearranging the equation, we get:

F_s = -ma_r - μkR

Substituting the known values, we get:

F_s = -(4 kg)(2.47 m/s2) - (0.3)(13.82 N) = -10.27 N

Therefore, the magnitude of the force exerted by the spring on the collar is approximately 10.27 N. It is important to note that the negative sign indicates that the force is acting in the opposite direction of the motion of the collar, which makes sense since the spring is trying to pull the collar towards point A while the collar is moving away from it.

In conclusion, the coefficient of dynamic friction plays a crucial role in determining the magnitude of the force exerted by the spring on the collar. It is also important to consider all the forces acting on the collar and use the appropriate equations to solve for the unknown variables.