Question

Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 = 1.2 kg and M2 = 2.8 kg. Block 1 rests a distance r = 0.45 m from the center of the turntable. The coefficient of static friction between block and turntable is μs = 0.54. Block 2 hangs vertically underneath.

Answer 1

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

$v = 3.5596 \frac{m}{s}$

Explanation:

The motion have the velocity that will make the block move using:

$F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}$

$m_{1} = 1.2 Kg$

$m_{2} = 2.8 Kg$

$r = 0,45 m$

μ$s= 0,54$

Resolving:

$\frac{m_{1}*v^{2} }{r} = us*m_{1}* g + m_{2} * g$

$v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }$

$v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}$

$v^{2} = 12.6714 \frac{m^{2} }{s^{2} }$

$v = 3.559 \frac{m}{s}$