Two blocks are connected by a light string passing over a pulley of radius 0.16
ID: 1262968 • Letter: T
Question
Two blocks are connected by a light string passing over a pulley of radius 0.16 m and moment of inertia I. The blocks move (towards the right) with an acceleration of 1.00 m/s2 along their frictionless inclines (see the figure).
Determine its moment of inertia, I.
Two blocks are connected by a light string passing over a pulley of radius 0.16 m and moment of inertia I. The blocks move (towards the right) with an acceleration of 1.00 m/s^2 along their frictionless inclines (see the figure). Determine its moment of inertia, I.Explanation / Answer
Net force on A, ?FA = FTA- m1g sin ?1 =m1.a
FTA =m1g sin ?1 + m1.a =m1(g sin ?1 + a)
= 8.0(9.8x0.5299 + 1) = 49.55 N
Similarly,
FTB= m2g sin ?2 -m2.a = m2(g sin ?2 -a)
= 10(9.8x0.8746 -1) = 75.71N
(c) Torque on the pulley = I.?
Angular acceleration,
? = a/R = 1.00/0.16= 6.25 rad/s2
Net force on the pulley,
?Fp = FTB - FTA = 75.71- 49.55 = 26.16 N
Net Torque on the pulley,
? = ?Fp .R= 26.16x0.16 =4.1856 J
Moment of inertia of the pulley,
I =?/? = 4.1856/6.25 = 0.67 kg.m2