A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer?

Question

A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity &lsquo;V&rsquo; towards the outer end of the rod and the rod is set to be rotating with an angular velocity &omega;. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)&omega; = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer?

Answer

For the ring to move in a circle at constant speed the net force on it should be zero. Here spring force will provide the necessary centripetal force.
∴ kx = mxω2

⇒

Answer

Ans is B Since radius will be contant so no effect no effect on velocity

Answer

For the ring to move in a circle at constant speed the net force on it should be zero. Here spring force will provide the necessary centripetal force.
∴ kx = mxω²

⇒

Answer

And the spring constant is 50 N/m. The natural length of the spring is 0.5m. The ring is initially at rest and is then displaced by 0.2m from its equilibrium position and released. Find the period of oscillation of the ring.

To find the period of oscillation, we can use Hooke's Law and the equation for the period of a mass-spring system.

Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as:

F = -kx

Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the spring constant is 50 N/m and the displacement is 0.2m. Plugging these values into the equation, we can find the force exerted by the spring:

F = -50 * 0.2
F = -10 N

Since the force is negative, it means that the force is acting in the opposite direction of the displacement, which is towards the equilibrium position.

Now, let's consider the forces acting on the ring. The only force acting on the ring is the force exerted by the spring. According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration:

F_net = m * a

Since the only force acting on the ring is the force exerted by the spring, we can say that F_net is equal to the force exerted by the spring:

F_net = -10 N

Since the ring is initially at rest, its initial velocity is zero. Therefore, the net force acting on the ring is equal to the mass of the ring multiplied by its acceleration:

-10 N = 3 kg * a

Solving for acceleration:

a = -10 N / 3 kg
a = -3.33 m/s^2

Again, the negative sign indicates that the acceleration is in the opposite direction of the displacement, which means it is towards the equilibrium position.

Now, let's use the equation for the period of a mass-spring system:

T = 2π * sqrt(m/k)

Where T is the period, m is the mass of the object, and k is the spring constant.

Plugging in the values:

T = 2π * sqrt(3 kg / 50 N/m)
T = 2π * sqrt(0.06 kg/N)
T ≈ 2π * 0.2449 s
T ≈ 1.54 s

Therefore, the period of oscillation of the ring is approximately 1.54 seconds.

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