Question

A 0.50 m long, uniform cylinder with a mass of 0.75 kg and radius 0.10 m is on a frictionless
axle. A light cord wrapped around its center runs over a pulley to a 1.40 kg hanging mass below.
The pulley is also on a frictionless axle and is a disc which has a mass of 0.20 kg and radius 0.05
m. If the system was held at rest and then released, how fast is the hanging mass moving after it
has fallen a total distance of 0.50 m? (Assume that the cord does not slip on either surface and is
always tight.)

Answer 1

Answer:

Explanation:

uniform cylinders and discs have moment of inertia of ½mR². This means that in a linear system, the cylinder or disc mass acts as if it has ½ of its actual value.

F = ma

1.4(9.8) = (1.4 + 0.75/2 + 0.20/2)a

a = 7.3173333... m/s²

v² = u² + 2as

v² = 0² + 2(7.3173)(0.50)

v = √7.3173 = 2.7050569...

v = 2.7 m/s

Answer 2

Hi there!

We can begin by identifying the moment of inertia for the objects that will rotate.

For both a uniform cylinder and disk, the moment of inertia is equivalent to:

$I = \frac{1}{2}MR^2$

Let:

T₁ = Tension of rope section connecting rolling cylinder to pulley

T₂ = Tension of rope section connecting hanging mass to pulley

m₁ = mass of cylinder

m₂ = mass of pulley

m₃ = mass of hanging block

a = acceleration of entire system

g = acceleration due to gravity

We can begin by doing a summation of torques about the pulley:

$\Sigma \tau = RT_2 - RT_1$

Rewrite using the rotational equivalent of Newton's Second Law:

$\Sigma \tau = I\alpha$

$I\alpha = RT_2 - RT_1$

Rewrite alpha as a/r and substitute in the moment of inertia:

$\frac{1}{2}M_2R^2\frac{a}{R} = RT_2 - RT_1$

Cancel out the radii:

$\frac{1}{2}M_2a = T_2 - T_1$

Now, we must solve for each tension.

T₁

Sum torques acting on mass 1 and use the same method as above:

$\Sigma \tau = RT_1$

$\frac{1}{2}M_1R^2\frac{a}{R} = RT_1$

$\frac{1}{2}M_1a = T1$

T₂

We can use a summation of forces:

$\Sigma F = W - T_2\\\Sigma F = m_3g - T_2\\T_2 = m_3g - m_3a$

Plug these derived expressions into the above:

$\frac{1}{2}M_2a = m_3g - m_3a - \frac{1}{2}M_1a$

Rearrange to solve for acceleration:

$a = \frac{m_3g}{\frac{1}{2}M_1+\frac{1}{2}M_2 + m_3}$

Solve for a:

$a = \frac{(1.40)(9.8)}{\frac{1}{2}(0.75)+\frac{1}{2}(0.2) + 1.4} = 7.317 m/s^2$

Now, we can use the following kinematic equation to solve for velocity given acceleration and distance:

$vf^2 = vi^2 + 2ad\\\\vf^2 = 2(7.317)(0.5)\\vf = \sqrt{2(7.317)(0.5)} = \boxed{2.705 m/s}$