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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.

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okpalawalter8

Answer:

The acceleration is  [tex]a = 5.88 \ m/s^2[/tex]

Explanation:

From the question we are told that

  The mass of the weight is  [tex]m_w = 150 \ g = 0.150 \ kg[/tex]

    The radius of the spool is  [tex]r = 4 \ cm = 0.04 \ m[/tex]

     The mass of the spool is [tex]m_c = 200 \ g = 0.20 \ kg[/tex]

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

     [tex]F_n = W - T[/tex]

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

     [tex]W = m_w *g[/tex]

and  T  is the tension on the thread

So

     [tex]F_n = m_w * g - T[/tex]

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

     [tex]F_n = m_w * a[/tex]

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

     [tex]m_w * a = m_w * g - T[/tex]

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

       [tex]\tau = T * r[/tex]

This torque is also mathematically represented as

      [tex]\tau = I * \alpha[/tex]

Here [tex]I[/tex] is moment of inertia of the spool which is mathematically represented as

       [tex]I = \frac{1}{2} * m_s * r^2[/tex]

while  [tex]\alpha[/tex] is the angular acceleration of the spool which is mathematically represented as

       [tex]\alpha = \frac{a}{ r}[/tex]

so

     [tex]\tau = \frac{1}{2} * m_s * r^2 * \frac{a}{r}[/tex]

=>  [tex]\tau =\frac{m_s * r * a}{2}[/tex]

So

     [tex]\frac{m_s * r * a}{2} = T * r[/tex]

=>   [tex]T = \frac{m_s * a}{2}[/tex]

Now substituting this formula for T  into the equation above  

        [tex]m_w * a = m_w * g - \frac{m_s * a}{2}[/tex]

=>     [tex]a = \frac{m_w * g}{m_w + \frac{m_s}{2} }[/tex]

=>     [tex]a = \frac{0.150 * 9.8}{0.150 + \frac{0.20}{2} }[/tex]

=>     [tex]a = 5.88 \ m/s^2[/tex]

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