PHY 2110 HW 22 -C | (i) www.webassign.net/web/StudentAssignment-Responses/submit

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PHY 2110 HW 22 -C | (i) www.webassign.net/web/StudentAssignment-Responses/submit?dep-17470511 -Professional NAS Computers nl Edit-Unsried Bookmarks MightyText B 4K Movies 4K Blur e Welcome l Granma r Cozy Hobbit Times, wr Wells Fargo Perer -Online Stock Tradin O Randy Pausch Last -/1 points SerPSET9 10.P.053. In the figure below the hanging object has a mass of m1-0.425 kg; the sliding block has a mass of m? 0.800 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m and an outer radius of R2-0.030 0 m. Assume the mass of the spokes is ne ligible. The coefficient af kinetic friction between the block and the horizontal surface ism-D.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table. My Notes Ask Your RI in (d) Use energy methods to predict its speed after it has moved to a second point, 0.700 m way m/s (b) Find the angular speed of the pulley at the same moment. rad/s Submit Answer Save Progress 5. -/1 points SerCP118.P059 My Notes Ask Your A 2.30 kg solid, uniform disk of radius 0.160 m is released from rest at point A in the figure below, its center of gravity a distance of 1.80 m above the ground. The disk rolls without slipping to the bottom of an incline and back up to point B where it is launched vertically into the air. The disk then rises to its maximum height hmx at point C. HINT |s+4 12:16

Explanation / Answer

Given,

m1 = 0.425 kg ; m2 = 0.8 kg ; M = 0.35 kg ;

R1 = 0.02 m ; R2 = 0.03 m ; uk = 0.25 ; vi = 0.82 m/s

A) Let V be this speed.

The change in KE would be equal to the net work done. Also, w = v/R2 and moment of inertia of given pulley will be = I = 1/2 M (R1^2 + R2^2)

1/2 ( m1 + m2) (V^2 - vi^2) + 1/2 I ( wf^2 - wi^2) = m1gh - uk m2g

1/2 ( m1 + m2) (V^2 - vi^2) + 1/2 x 1/2 M (R1^2 + R2^2) x (V^2 - vi^2) / R2^2 = m1gh - uk m2g

Solving this for V we get

V = sqrt { vi^2 + [ ( m1gh - uk m2g ) / 1/2 (m1 + m2) + 1/2 M ( 1 + R1^2/R2^2) ] }

Putting in the values we get

V = sqrt { (0.82)^2 + [ ( 0.425 x 9.81 x 0.7 - 0.25 x 0.8 x 9.81 ) / [0.5 x (0.425 + 0.8) + 0.5 x 0.35 x (1 + 0.02^2/0.03^2)]} = 1.15

V = sqrt [ 0.6724 + 0.4484/0.8453 ] = 1.15 m/s

Hence, V = 1.15 m/s

b) w = V/R2 = 1.15/0.03 = 38.33 rad/s

Hence, w = 38.33 rad/s

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