# Rotational motion and conservation of energy problem

• haydn
In summary, the problem involves a sliding block with a mass of 0.800 kg, a counterweight with a mass of 0.460 kg, and a pulley with a mass of 0.350 kg and an inner radius of 0.020 m and an outer radius of 0.030 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle and the light cord does not stretch or slip on the pulley. The block has an initial velocity of 0.820 m/s and is moving towards a photogate. The question is to use energy methods to predict its speed after it has moved to a second photogate

## Homework Statement

The sliding block has a mass of 0.800 kg, the counterweight has a mass of 0.460 kg, and the pulley is a hollow cylinder with a mass of 0.350 kg, an inner radius of 0.020 m, and an outer radius of 0.030 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward the pulley when it passes through a photogate.

(a) Use energy methods to predict its speed after it has moved to a second photogate, 0.700 m away.

## Homework Equations

KET = (1/2)mv2
KER = (1/2)Iw2

I for a hollow cylinder = (1/2)M(R12 + R22)

w = omega = angular velocity

## The Attempt at a Solution

I know work is equal to the change in kinetic energy so I go:

W = ((1/2)(M1+M2)Vf2 + (1/2)IwF2) - ((1/2)(M1 + M2)VI2 + (1/2)Iw2 + M2gs)

and W = -f * s = - (coefficient of friction)M1gs

s = .7 as given in the problem...

So I substitute that expression for work into the first equation, solve for VF but the website I'm using is telling me I'm getting the wrong answer... what am I doing wrong?

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haydn said:
and W = -f * s = - (coefficient of friction)M1gs

That's not right. There is no reason that the work done by friction must equal the change in gravitational potential energy of the object. The Work-Energy Theorem states that the net work done on a system equals the change in total kinetic energy of the system. So we have:

$$W_{total}=\Delta K_{total}$$

$$W_f+W_g=\Delta K_{block}+\Delta K_{pulley}+\Delta K_{counterweight}$$

$$-Fs+m_2gs=\frac{1}{2}m_1\left(v_2^2-v_1^2\right)+\frac{1}{2}I\left(\omega_2^2-\omega_1^2\right)+\frac{1}{2}m_2\left(v_2^2-v_1^2\right)$$

You'll need to relate $\omega$ to $v$, but other than that this should be a piece of cake. You know everything except $v_2$.

Ok, isn't friction force equal to the coefficient of friction times the normal force though? ..and the normal force of the block = M1g since it isn't moving vertically?

Thanks for the help by the way, besides that everything is clear now.

haydn said:
Ok, isn't friction force equal to the coefficient of friction times the normal force though? ..and the normal force of the block = M1g since it isn't moving vertically?

Yes on both counts.

Thanks for the help by the way, besides that everything is clear now.

No problem!

## 1. What is rotational motion and how is it different from linear motion?

Rotational motion is the movement of an object around an axis or point, while linear motion is the movement of an object in a straight line. In rotational motion, the object's position changes along a curved path, while in linear motion, the object's position changes along a straight path.

## 2. What is conservation of energy and how does it relate to rotational motion?

Conservation of energy is the principle that energy cannot be created or destroyed, only transferred from one form to another. In rotational motion, this means that the total kinetic and potential energy of a system remains constant, as long as there are no external forces acting on the system.

## 3. How do you calculate the rotational kinetic energy of an object?

The rotational kinetic energy of an object is calculated using the formula KE = 1/2Iω^2, where I is the moment of inertia and ω is the angular velocity. Moment of inertia is a measure of an object's resistance to rotational motion.

## 4. Can rotational motion be used to do work?

Yes, rotational motion can be used to do work. For example, when a force is applied to a wrench to loosen a bolt, the rotational motion of the wrench is doing work on the bolt by applying a torque.

## 5. How does the conservation of energy apply to a pendulum?

In a pendulum, the potential energy is converted to kinetic energy and back again as the pendulum swings back and forth. The total energy (potential + kinetic) remains constant, demonstrating the principle of conservation of energy.