The power of a bus engine is transmitted using the beltandpu

The power of a bus engine is transmitted using the belt-and-pulley arrangement shown. The engine turns pulley A at ?A=(20t+35)rad/s clockwise, where t is in seconds. (Figure 1)

Part A

Determine the angular velocity of the generator pulley B, measured clockwise, when t = 3 s.

Express your answer using three significant figures. Enter positive value if the angular velocity is clockwise and negative value if the angular velocity is counterclockwise.

The power of a bus engine is transmitted using the belt-and-pulley arrangement shown. The engine turns pulley A at ?A=(20t+35)rad/s clockwise, where t is in seconds. (Figure 1)

Part A Determine the angular velocity of the generator pulley B, measured clockwise, when t = 3 s. Express your answer using three significant figures. Enter positive value if the angular velocity is clockwise and negative value if the angular velocity is counterclockwise.

session.masteringengineering.comm EGM 202 81-83 Signed in as khaled almuaibed Help Close Assignment #11 Problem 16.12 Resources us 1 of 2 | next » Problem 16.12 Part A The power of a bus engine is transmitted using the belt and-pulley arrangement shown. The engine turns pulley A at wA (20t +35) rad/s clockwise, where t is in seconds. (Figure 1) Determine the angular velocity of the generator pulley B, measured clockwise, when t 3 s Express your answer using three significant figures. Enter positive value if the angular velocity is clockwise and negative value if the angular velocity is counterclockwise vec rad/s Submit My Answers Give Up Part B Determine the angular velocity of the air-conditioning pulley C, measured clockwise, when t 3 s Figure 1 of 1 Express your answer using three significant figures. Enter positive value if the angular velocity is clockwise and negative value if the angular velocity is counterclockwise 100 mm vec rad/s 75 mm Submit My Answers Give Up 50 mm Provide Feedback Continue

Solution

When t = 3 sec
The angular velocity of A = 20x3 + 35 = 95 rad/sec

clearly, D = B ,, since they are mounted on same shaft

Considering the fact that the velocity of belt will remain constant for the entire body of the belt. therefore the velocity of belt on the tip of pulley A must be same as the velocity of the belt on the tip of pully D

which implies , (r)_A =  (r)_D

hence we get :

D /A = rA / rD = 75 / 25 = 3

D = A *3 = 95*3 = 285 rad / sec

B =D = 285 rad/sec

Now, C / B = rB / rC = 100 / 50 = 2

C = B x 2 = 285*2=570 rad / sec