Hi there!
1.
Let's use the rotational equivalent of Newton's Second Law.
[tex]\Sigma \tau = I\alpha[/tex]
τ = Torque (Nm)
I = Moment of Inertia (1/2mR² for solid cylinder)
α = Angular acceleration (rad/sec²)
The torque is equivalent to:
[tex]\tau = r \times F[/tex]
r = distance from lever arm to pivot point, aka the radius for this cylinder (0.111 m)
F = applied force (7.161 N)
We are already given various values, so let's create a working equation and solve.
[tex]rF = I\alpha \\\\\alpha = \frac{rF}{I}[/tex]
[tex]\alpha = \frac{(0.111)(7.161)}{\frac{1}{2}(2.15)(0.111^2)} = \boxed{60.013 \frac{rad}{sec^2}}[/tex]
2.
So, let's begin by doing a summation of forces acting on the block.
We have its force of gravity downward(+), and the tension from the string upward. (-), assigning signs based on the path of the falling block.
[tex]\Sigma F = F_g - T\\\\\Sigma F = m_b g - T\\\\m_b a = m_b g - T[/tex]
Solving for 'T':
[tex]T = m_b g - m_b a[/tex]
Let's now do a summation of torques on the cylinder. We only have the torque caused by the tension in the string connected to the block.
[tex]\Sigma \tau = rT\\\\I\alpha = r(T = m_b g - m_b a)[/tex]
Using the relationship between angular and translational acceleration:
[tex]a = \alpha r\\\\alpha = \frac{a}{r}[/tex]
And the equation for the moment of inertia:
[tex]I = \frac{1}{2}Mr^2[/tex]
Simplify the expression.
[tex]\frac{1}{2}Mr^2(\frac{a}{r}) = r( m_b g - m_b a)\\\\\frac{1}{2}Ma = ( m_b g - m_b a)[/tex]
Solve for 'a':
[tex]\frac{1}{2}ma + m_b a= m_b g \\\\a = \frac{m_b g}{\frac{1}{2}M + m_b} = \frac{0.730(9.81)}{\frac{1}{2}(2.15) + 0.730} = \boxed{3.97 \frac{m}{s^2}}[/tex]
3.
This is just a case of kinematics since we have a constant acceleration. To solve for this instance, however, let's use calculus.
We know that:
[tex]a(t) = 3.97[/tex]
v(t) is the integral. There is no initial velocity (block starts from rest), so:
[tex]v(t) = 3.97t + C\\C = 0 m/s \\\\v(t) = 3.97t[/tex]
Now, we can take the integral of the velocity/time equation to find the displacement between the interval.
[tex]\int\limits^{0.590}_{0.390} {3.97t} \, dt = \boxed{0.389 m}[/tex]
4.
So, let's now find the translational acceleration produced by the new cylinder. Use the kinematic equation:
[tex]\Delta x = v_0 t + \frac{1}{2}at^2[/tex]
Since the block starts from rest, we can do some simplifying and rearranging:
[tex]2\Delta x = at^2\\\\a = \frac{2\Delta x}{t^2} = \frac{2(0.450)}{0.490^2} = 3.748 \frac{m}{s}^2[/tex]
So, let's go back to our summation of torques equation. This time, however, we cannot simplify 'I'.
[tex]I(\frac{a}{r}) = r( m_b g - m_b a)[/tex]
Solving for 'I':
[tex]I = \frac{r^2( m_b g - m_b a)}{a} = \frac{(0.111^2)(7.161 - .730(3.748))}{3.748}\\\\ = \boxed{0.0145 kgm^2}[/tex]
