- #1

JD_PM

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## Homework Statement

You buy a bottle of water in the store and place it on the conveyor belt with the longitudinal axis perpendicular to the direction of movement of the belt. Initially, both the belt and the bottle are at rest. We can approach the bottle as one cylinder with radius ##R##, mass ##M## and moment of inertia ##I = M R^2##, in which the mass is not distributed uniformly. The speed of the belt at time ##t## is ##V (t)##.

a) Find an expression for the speed ##v(t)## of the centre of mass of the bottle.

## Homework Equations

Second Newton's Law for rotation:

$$\tau = I \alpha = Rf$$

## The Attempt at a Solution

I got two expression for the speed of the centre of mass of the bottle, and I wanted to know if both are equivalent and correct.[/B]

Using the second Newton's Law for rotation you end up with the expression:

EQ1:

$$R\alpha=2a_0$$

Additionally, the total acceleration is the sum of the tangential and centripetal vector components of the acceleration. Thus:

EQ2:

$$a = a_o + R\alpha$$

With this info we can deduce that:

EQ3:

$$a_0=\frac13 a$$

EQ4:

$$R\alpha=\frac23 a$$

EQ5:

$$f=\frac13 Ma$$

Therefore,

**using the second Newton's Law for rotation I got the following speed:**

EQ6:

$$v_{CM} = \frac{3R \omega}{2}$$

**And using kinematics:**

$$v_f^2 = v_o^2 + 2a(x_f - x_o)$$

I got:

$$v_{CM} = \sqrt{ \frac{6fd}{M}}$$

Note that ##d## is the distance the rolling cylinder covers.

**Are these correct and equivalent?**

Thank you