Let the masses of blocks A and B be 4.50 kg and 2.00 kg, respectively, the momen

Question

Let the masses of blocks A and B be 4.50 kg and 2.00 kg, respectively, the moment of inertia of the wheel about its axis be 0.300 kg*m2 and the radius of the wheel be 0.100 m.

1. Find the linear accelerations of block A if there is no slipping between the cord and the surface of the wheel.

2. Find the linear accelerations of block B if there is no slipping between the cord and the surface of the wheel.

4.Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel.

5.Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.

Lengthy - Please, please...help with all parts and show work - really appreciate it!

Explanation / Answer

T1 is tension in A
T2 is tension in B
(T1-T2)r=I

there is no slipping so

a=r

(T1-T2)r=Ia/r

m1g-T1=m1a

T2-m2g=m2a

solving these 3 equation 3 variables we get

1.a=0.672m/s^2 downwards

2.a=0.672m/s^2 upwards

3.=a/r=6.72 rad/s^2

4.T2=20.96 N

5.T1=41.12 N

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