Respuesta :
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]