Question

A weight with mass mw=150 gmw=150 g is tied to a piece of thread wrapped around a spool, which is suspended in such a way that it can rotate freely. When the weight is released, it accelerates toward the floor as the thread unwinds. Assume that the spool can be treated as a uniform solid cylinder of radius R=4.00 cmR=4.00 cm and mass ????s=200 gMs=200 g . Find the magnitude aa of the acceleration of the weight as it descends. Assume the thread has negligible mass and does not slip or stretch as it unwinds.

Answer 1

The acceleration is
`a = 5.88 \ m/s^2`

Step-by-step explanation:

From the question we are told that

The mass of the weight is
`m_w = 150 \ g = 0.150 \ kg`

The radius of the spool is
`r = 4 \ cm = 0.04 \ m`

The mass of the spool is
`m_c = 200 \ g = 0.20 \ kg`

Generally the net force acting on the weight is mathematically represented as

`F_n = W - T`

Here W is the weight of the weight which is mathematically represented as

`W = m_w *g`

and T is the tension on the thread

So

`F_n = m_w * g - T`

Generally this net force acting on the weight can be mathematically represented as

`F_n = m_w * a`

Here is the a is the acceleration of the system (i.e acceleration of the weight as a result of its weight and the tension on the rope )

So

`m_w * a = m_w * g - T`

Generally the torque which the spool experiences can be mathematically represented as

`\tau = T * r`

This torque is also mathematically represented as

`\tau = I * \alpha`

Here
`I` is moment of inertia of the spool which is mathematically represented as

`I = (1)/(2) * m_s * r^2`

while
`\alpha` is the angular acceleration of the spool which is mathematically represented as

`\alpha = (a)/( r)`

so

`\tau = (1)/(2) * m_s * r^2 * (a)/(r)`

=>
`\tau =(m_s * r * a)/(2)`

So

`(m_s * r * a)/(2) = T * r`

=>
`T = (m_s * a)/(2)`

Now substituting this formula for T into the equation above

`m_w * a = m_w * g - (m_s * a)/(2)`

=>
`a = (m_w * g)/(m_w + (m_s)/(2) )`

=>
`a = (0.150 * 9.8)/(0.150 + (0.20)/(2) )`

=>
`a = 5.88 \ m/s^2`