Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 334: 10.63

(a) $a = 2.88~m/s^2$ (b) $a = 6.13~m/s^2$ A lower tension is required in the string to make the disks rotate, and, therefore, the block has a higher acceleration.

(a) We can find the moment of inertia of the two disks. $I = \frac{1}{2}M_1~R_1^2+\frac{1}{2}M_2~R_2^2$ $I = \frac{1}{2}(0.80~kg)(0.0250~m)^2+\frac{1}{2}(1.60~kg)(0.0500~m)^2$ $I = 0.00225~kg~m^2$ We can set up a torque equation for the disks. $\tau = I\alpha$ $T~R_1 = I(\frac{a}{R_1})$ $T = \frac{Ia}{R_1^2}$ We can use this expression in the force equation for the block. Let $m$ be the mass of the block. $\sum F = ma$ $mg-T = ma$ $mg = ma+T$ $mg = ma+ \frac{Ia}{R_1^2}$ $a = \frac{mg}{m+ \frac{I}{R_1^2}}$ $a = \frac{(1.50~kg)(9.80~m/s^2)}{1.50~kg+ \frac{0.00225~kg~m^2}{(0.0250~m)^2}}$ $a = 2.88~m/s^2$ (b) We can set up a torque equation for the disks. $\tau = I\alpha$ $T~R_2 = I(\frac{a}{R_2})$ $T = \frac{Ia}{R_2^2}$ We can use this expression in the force equation for the block. Let $m$ be the mass of the block. $\sum F = ma$ $mg-T = ma$ $mg = ma+T$ $mg = ma+ \frac{Ia}{R_2^2}$ $a = \frac{mg}{m+ \frac{I}{R_2^2}}$ $a = \frac{(1.50~kg)(9.80~m/s^2)}{1.50~kg+ \frac{0.00225~kg~m^2}{(0.0500~m)^2}}$ $a = 6.13~m/s^2$ The acceleration is greater when the string is wrapped around the larger disk. This makes sense because there will be more torque on the disks for a given force when the string is wrapped around a larger radius. This means that a lower tension is required in the string to make the disks rotate, and, therefore, the block has a higher acceleration.