Circular motion and energy of a massless spring

• nomorenomore
In summary, the minimum speed needed for the mass at point C to make it through the loop-the-loop is vC.
nomorenomore

Homework Statement

A massless spring of constant k is fixed on the left side of a level track. A block of mass m is pressed against the spring and compresses it a distance d, as shown in the figure. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. Given that the coefficient of kinetic friction between the block and the track along AB is k, and that the length of AB is L. The gravitational acceleration is g.

a) Express the minimum speed of the mass at point C, vC, in terms of symbols given such that it can make through the loop-the-loop (i.e. remain in contact with the track).
(b) Hence or otherwise, determine the minimum compression, d, of the spring for such situation.

Homework Equations

Fnet=ma,
ac=v^2/R,
Total energy = KE + PE + friction
Spring energy = -kx

The Attempt at a Solution

a) Fnet=ma
N+mg=m(ac)
2mg=m(v^2/R)
2g=(v^2/R)
vc=sqt(2gR)

b) -kd=1/2m(vc)^2 + mgR + k
d= (1/2m(vc)^2 + mgR + k )/-k

Are these the correct approaches, please?

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In (a), where does 2 in 2mg come from?

In (b), what does the last term mean?

1 person
voko said:
In (a), where does 2 in 2mg come from?

In (b), what does the last term mean?

In (a), i thought N + mg = m(ac) where N=mg
therefore mg + mg = 2mg = m(ac)
So... maybe I'm wrong?

In (b), -kd=1/2m(vc)^2 + mgR + (μk)*mg*L
d= (1/2m(vc)^2 + mgR + (μk)*mg*L )/-k

In (a), why would N = mg?

In (b), the right hand side has the dimension of energy. The left hand side, however, has the dimension of force. That cannot be correct.

voko said:
In (a), why would N = mg?

In (b), the right hand side has the dimension of energy. The left hand side, however, has the dimension of force. That cannot be correct.

In (a), because I thought N is always equal to mg...
But just now I've thought it twice.. it's not true, a simple example is the Lift Case..
So N > 0 ; mg > m(ac) ; g > v^2/r ; v > sqrt(g*r) ?

In (b), for both sides having the dimension of energy, I should use 1/2*k*x^2 instead of -kd then??

thanks

(b) is correct now.

For (a), observe that you are asked about the minimum speed that would still keep the block in contact with track. Which means that if the speed is a just tiny bit less, it will just fall down. Would there be any reaction force from the track when the block is just short of losing contact with it?

1. What is circular motion?

Circular motion is the movement of an object along a circular path, where the object maintains a constant distance from a fixed point.

2. How does a massless spring work?

A massless spring is a theoretical concept used in physics to simplify calculations. It is a spring with no mass and no friction, meaning that it can stretch and compress infinitely without losing any energy.

3. Can a massless spring have energy?

Yes, a massless spring can have energy in the form of potential energy. When it is stretched or compressed, it stores potential energy that can be released when the spring returns to its original shape.

4. How is the energy of a massless spring related to circular motion?

In circular motion, the energy of a massless spring is constantly changing between kinetic energy and potential energy. As the object moves along the circular path, the spring stretches and compresses, converting kinetic energy into potential energy and vice versa.

5. What factors affect the circular motion and energy of a massless spring?

The circular motion and energy of a massless spring can be affected by the mass and velocity of the object, the stiffness of the spring, and the radius of the circular path. Any changes in these factors can alter the amount of energy being converted and the overall motion of the object.

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