Two metal disks, one with radius R_1 = 2.50 cm and mass M_1 = 0.80 kg and the ot

Question

Two metal disks, one with radius R_1 = 2.50 cm and mass M_1 = 0.80 kg and the other with radius R_2 = 5.00 cm and mass M_2 = 1.60 kg, are welded together and mounted on a frictionless axis through their common center (Fig. P9.87). What is the total moment of inertia of the two disks? A light siring is wrapped around the edge of the smaller disk, and a 1.50-kg block is suspended from the free end of the string. If the block is released from rest at a distance of 2.00 m above the floor, what is its speed just before it strikes the floor? Repeat the calculation of pan (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain why this b so.

Explanation / Answer

total MOI of system = m1R1^2 / 2 + M2 * R2^2 /2

=0.80*(2.5/100)^2 / 2 +1.60* (5/100)^2 /2

=0.00225 Kgm2

after 1.50 Kg block is attached to it

torque = 1.50 * 9.81 *R1 =1.50*9.81*(2.50/100)

angular acc = torque / I => (1.50*9.81*(2.50/100))/(0.00225) = 163.5

linear acceleration = alpha * R1 => 163.5*(2.5/100) =4.0875 m/s2

using NLM

2 = 0+ 0.5*4.0875 * t^2

t= sqrt (2/(0.5*4.0875))

=0.98923873365 sec

V= u +at

= 4.0875*0.98923873365

=4.04351332379 m/s

if larger disk is used

angular acc= torque / I => ( 163.5*2) { it is doubled as torque is doubled}

now linear acceleration = alpha * R2 => (4*4.0875)

clearly acceleration is four times initial acc

t= 0.49461936682 sec

V= u +at

= 4.0875*4*0.49461936682

=8.09 m/s

velocity is doubled as torque was increased since torqure is dependent on distance from axis of rotation

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