4. An object with a mass of m 5.20 kg is attached to the free end of a light str

Question

4. An object with a mass of m 5.20 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.350 m and mass M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about its center as shown in Figure 2. The suspended mass is released from rest 6.50 m above the floor. Determine (a) the tension in the string, (b the acceleration of the object, and (c) the speed with which the object hits the floor. (d) Verify this velocity found in (c) using conservation of energy. (1300 ml and mass·l =

Explanation / Answer

From conservation of energy

PE = KE(rot) + KE(travs)

m g h = 1/2 m v^2 + 1/2 I w^2

I = M R^2/2 ; w = v/R

m g h = 1/2 m v^2 + 1/2 (1/2 M R^2)(v/R)^2

m g h = 1/2 v^2 (m + M/2)

1/2 v^2 = m g h / (m + M/2)

v = sqrt [ 4 m g h/ 2m + M ]

v = sqrt [4 x 5.2 x 9.8 x 6.5/ (2 x 5.2 + 3)] = 9.94 m/s

Hence, v = 9.94 m/s ----------> Ans(c)

from eqn of motion we know

v^2 = u^2 + 2 a s

In our case, u = 0 ; s = h ; v = sqrt [ 4 m g h/ 2m + M ] ;

[sqrt (4 m g h/ 2m + M)]^2 = 2 a h

a = 2 m g/ 2m + M

a = 2 x 5.2 x 9.8/(2 x 5.2 + 3) = 7.61 m/s^2

Hence, a = 7.61 m/s^2 -----------------> Ans(b)

Let T be the tension in the string, so for the net force we can write

ma = mg - T

T = m (g - a)

T = 5.2 (9.8 - 7.61) = 11.4 N

Hence, T = 11.4 N ----------------------------. Ans(a)

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