Two masses are suspended from a pulley as shown in the figure ( Figure 1 ) . The

Question

Two masses are suspended from a pulley as shown in the figure (Figure 1) . The pulley itself has a mass of 0.40 kg , a radius of 0.20 m , and a constant torque of 0.25 mN due to the friction between the rotating pulley and its axle. Two masses are suspended from a pulley as shown in the figure (Figure 1) . The pulley itself has a mass of 0.40 kg , a radius of 0.20 m , and a constant torque of 0.25 mN due to the friction between the rotating pulley and its axle. Two masses are suspended from a pulley as shown in the figure (Figure 1) . The pulley itself has a mass of 0.40 kg , a radius of 0.20 m , and a constant torque of 0.25 mN due to the friction between the rotating pulley and its axle. Two masses are suspended from a pulley as shown in the figure (Figure 1) . The pulley itself has a mass of 0.40 kg , a radius of 0.20 m , and a constant torque of 0.25 mN due to the friction between the rotating pulley and its axle.

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Provide FeedbackContinue

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining a= m/s2 a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Incorrect; Try Again; 3 attempts remaining Incorrect; Try Again; 3 attempts remaining Provide FeedbackContinue Figure 1 of 1 Two masses are suspended from a pulley as shown in the figure (Figure 1) . The pulley itself has a mass of 0.40 kg , a radius of 0.20 m , and a constant torque of 0.25 mN due to the friction between the rotating pulley and its axle.

Part A

What is the magnitude of the acceleration of the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ? (Neglect the mass of the string.) Express your answer using two significant figures. a= m/s2 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Provide FeedbackContinue Figure 1 of 1 T2 12 Im2g

Explanation / Answer

Moment of inertia of pulley, I=MR2/2 = 0.4*0.22/2 = 8*10-3 kg-m2

Let angular acceleration be A,then acceleration, a of the masses is

a= A*R,

A=a/0.2 =5a

net torque,T=0.8g -0.4g -0.25

T= 3.67 N-m

Now ,T =(0.8+0.4)a + IA

=1.2*a + 0.008*5a

hence

3.67=1.2a+0.04a

3.67=1.24 a

a=2.959 m/s2

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