In the R.-0-020 0 m, and an outer radius of R2-0.030 0 m. Assume pulley turns wi

Question

In the R.-0-020 0 m, and an outer radius of R2-0.030 0 m. Assume pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a table. 0.490 kg: the siding block has a mass of m 0.815 kg: and the puiley is a hollew cylinder with a mass of M-0.350 kg, an inner radus o the mass of the spokes is negligible. The coefficient of kinetic friction between velocity of v-0.820 m/s toward the pulley when the block and the horizontal surface is? . 0.250. The it pesses a reference paint on the (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away m/s (b) Find the angular speed of the pulley at the same moment (a) If they are released from rest and roll without slipping, which object reaches the bottom first? side by side at the top of an incline of height h.

Explanation / Answer

from the law of conservation of energy we have

(0.5 m2 v^2 - 0.5 m2 v1^2) + (0.5 m1 v^2 - 0.5 m1 v1^2) + (0.5 IW^2 - 0.5 I W1^2) + m1g (y-y1) + fk d = 0

Solving for V we get final formula as

V^2 = v1^2 + 4gd(m1-um2)/(2(m1+m2) + M(1 + (R1/R2)^2))

here

v1 = 0.82 m/s

g = 9.81 m/s^2

m1 = 0.49 kg

m2 = 0.815 kg

u = 0.25

R1 = 0.020 m

R2 = 0.03 m

d = 0.7 m

so

V^2 = 0.82^2 + 4* 9.81* 0.7*(0.49- 0.25 *0.815)/(2*(0.49+ 0.815) + 0.35 *(1 + (0.02/0.03)^2))

V = 1.78 m/s

-------

for using V = r W

W = angular velocity = V/r

W = 1.78/0.03

W = 54.6 rad/s

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