h=at²/2
a=2h/t²=2•0.76/5.2²=0.056 m/s²
Two equations for the loads
-m₁a=m₁g-T₁ => T₁=m₁(g+a) ….(1)
m₂a=m₂g-T₂,=> T₂=m₂ (g+a) ….(2)
and the 3rd equation (on the base of Newton’s 2 law for rotation Iε=M)
I•a/R=( T₂-T₁)R …. (3)
Substitute (1) and (2) in (3)
and solve for I
block 1 has mass m1 = 480 g, block 2 has mass m2 = 540 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.2 cm. When released from rest, block 2 falls 76 cm in 5.2 s (without the cord slipping on the pulley). What is the pulley's rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.
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