Question

A solid uniform 3.25-kg cylinder, 50.0 cm in diameter and 12.4 cm long, is connected to a 1.50-kg weight over two massless frictionless pulleys. The cylinder is free to rotate about an axle through its center perpendicular to its circular faces, and the system is released from rest. How far must the 1.50- kg weight fall before it reaches a speed of 2.10 m/s ? How fast is the cylinder turning at this instant?

Answer 1

0.469 m

8.40 rad/s

Step-by-step explanation:

Energy is conserved:

Potential energy = kinetic energy + rotational energy

PE = KE + RE

mgh = ½ mv² + ½ Iω²

Assuming the string doesn't stretch, the speed of the weight is equal to the tangential velocity of the cylinder, so ω = v/r.

For a cylinder, the moment of inertia about its axis is I = ½ Mr².

Substituting:

mgh = ½ mv² + ½ (½ Mr²) (v/r)²

mgh = ½ mv² + ¼ Mv²

Plugging in values:

(1.50) (9.8) h = ½ (1.50) (2.10)² + ¼ (3.25) (2.10)²

h = 0.469 m

The angular velocity of the cylinder is:

ω = v/r

ω = (2.10 m/s) / (0.250 m)

ω = 8.40 rad/s

Answer 2

This problem, we can use the conservation of energy principle. The potential energy lost by the falling weight is converted into the kinetic energy of both the falling weight and the rotating cylinder.

Let's denote:

`\( m_1 \)`: mass of the falling weight (1.50 kg)

`\( m_2 \)`: mass of the cylinder (3.25 kg)

R : radius of the cylinder (0.25 m, half of the diameter)

h : distance the weight falls

g : acceleration due to gravity
`(approximately \(9.8 \ \text{m/s}^2\))`

The potential energy lost by the falling weight is converted into the kinetic energy of both the falling weight and the rotating cylinder:

`\[ m_1gh = (1)/(2) m_1v^2 + (1)/(2) I\omega^2 \]`

Where:

`\( v \)` is the speed of the falling weight (2.10 m/s)

`\( I \)` is the moment of inertia of the cylinder about its central axis

`\( \omega \)` is the angular velocity of the cylinder

The moment of inertia
`\( I \)` for a solid cylinder about its central axis is
`\( I = (1)/(2) m_2R^2 \).`

Now, let's substitute the values and solve for
`\( h \)` and
`\( \omega \)`:

`\[ 1.50 \cdot 9.8 \cdot h = (1)/(2) \cdot 1.50 \cdot (2.10)^2 + (1)/(2) \cdot (1)/(2) \cdot 3.25 \cdot (0.25)^2 \cdot \omega^2 \]`

Solving for
`\( h \):`

`\[ 14.7h = 1.575 + 0.1015625\omega^2 \]`

Now, we need to find
`\( \omega \)`. The linear velocity of a point on the rim of the cylinder is given by
`\( v_{\text{linear}} = R\omega \)`. Since
`\( v_{\text{linear}} \)` is also the velocity of the falling weight, we have:

`\[ 2.10 = 0.25\omega \]`

Solving for
`\( \omega \):`

`\[ \omega = (2.10)/(0.25) = 8.4 \ \text{rad/s} \]`

Now, substitute this value back into the equation for
`\( h \)`:

`\[ 14.7h = 1.575 + 0.1015625 \cdot (8.4)^2 \]`

`\[ 14.7h = 1.575 + 7.57425 \]`

`\[ 14.7h = 9.14925 \]`

`\[ h \approx 0.623 \ \text{m} \]`

So, the weight must fall approximately 0.623 meters before it reaches a speed of 2.10 m/s. Now, to find the angular velocity
`\( \omega \)` of the cylinder:

`\[ \omega = (2.10)/(0.25) = 8.4 \ \text{rad/s} \]`

Therefore, the cylinder is turning at an angular velocity of 8.4 rad/s at this instant.