Answer
(a) $T = 42.0~N$
(b) $v = 11.8~m/s$
(c) $t = 1.69~s$
(d) $F = 160~N$
Work Step by Step
(a) We can set up an equation for the cylinder. Let $m_c$ be the mass of the cylinder.
$\tau = I\alpha$
$TR = \frac{1}{2}m_cR^2(\frac{a}{R})$
$a = \frac{2T}{m_c}$
We can use this expression in the force equation for the bucket. Let $m_b$ be the mass of the bucket.
$m_bg-T = m_ba$
$m_bg-T = m_b(\frac{2T}{m_c})$
$m_c~m_bg-m_cT = 2m_bT$
$T = \frac{m_c~m_bg}{2m_b+m_c}$
$T = \frac{(15.0~kg)(12.0~kg)(9.80~m/s^2)}{(2)(15.0~kg)+12.0~kg}$
$T = 42.0~N$
(b) We can find the acceleration.
$a = \frac{2T}{m_c}$
$a = \frac{(2)(42.0~N)}{12.0~kg}$
$a = 7.00~m/s^2$
We can find the speed after the bucket falls 10.0 meters.
$v^2=v_0^2+2ay = 0 +2ay$
$v = \sqrt{2ay} = \sqrt{(2)(7.00~m/s^2)(10.0~m)}$
$v = 11.8~m/s$
(c) We can find the time to fall 10.0 meters.
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}}$
$t = \sqrt{\frac{(2)(10.0~m)}{7.00~m/s^2}}$
$t = 1.69~s$
(d) The upward force exerted by the axle on the cylinder is equal in magnitude to the sum of the cylinder's weight and the tension pulling downward on the cylinder.
$F = m_cg+T$
$F = (12.0~kg)(9.80~m/s^2)+42.0~N$
$F = 160~N$
