An object with a mass of m = 4.6 kg is attached to the free end of a light strin

Question

An object with a mass of m = 4.6 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.255 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 5.00 m above the floor.

(a) Determine the tension in the string.
N

(b) Determine the magnitude of the acceleration of the object.
m/s2

(c) Determine the speed with which the object hits the floor.
m/s

(d) Verify your answer to part (c) by using the isolated system (energy) model. (Do this on paper. Your instructor may ask you to turn in this work.)

Explanation / Answer

angular velocity of reel be w and velocity of the mass be v

then v = Rw

and also acceleration a = R*alpha

force balance equation along vertical direction gives us

mg - T = ma

and Torque = I*alpha

T*R = MR^2/2 *alpha

(mg-ma) = MR/2 * a/R = Ma/2

solving this we get a = 7.4 m/s^2

a)

Tension = mg - ma = 11.04 N

b)

a = 7.4 m/s^2

c)

v^2 - u^2 = 2as

v = sqrt(2as) = sqrt(2*7.4*5) = 8.6 m/s

d)

using conservation of energy model

loss in potential energy of m = gain in KE of (m+M)

mgh = .5mv^2 + .5Iw^2

mgh = .5mv^2 + MR^2 * v^2/4R^2

mgh = .5mv^2 + Mv^2 / 4

solving thie we get v = 8.6

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.