Question

If we change the axis to ?2 ball as shown. Determine the moment of inertia about the given axis of rotation. Calculate the torque magnitude acting on the system.

nau be the speed if no downforce acted on the car? The drawing shows a system of objects, which consists of three small balls connected by massless rods. The axis is perpendicular to the page as shown. The force of magnitude F is applied to the m2 ball (see the drawing). The masses of the balls are mi = 9.00 kg, m2=6.00 kg, and m3= 7.00 kg. The magnitude of the force is F=424 N. F F mi mi 3.00 m 3.00 m OM2 m2 Axis 4.00 m 4.00 m 5.00 m 5.00 m Axis m3 m3 System A System B a. Determine the moment of inertia about the given axis of rotation. b. Calculate the torque magnitude acting on the system. c. The systems start from rest, and the direction of the force moves with the system and always points along the 4.00-m rod. What is the angular velocity of the system after 5.00 s? d. If we change the axis to m2 ball as shown. Determine the moment of inertia about the given axis of rotation. Calculate the torque magnitude acting on the system. A 2 kg block is hanging from a massless cord that is wrapped around a pulley I = 1.1 x 10-8kg m², as the nuounted from rotating and the block is stationary. Then, the pulley is dum Till

Answer 1

Note: Please look at part d for your answer.
a. The moment of inertia for each system is System A 1 = m,'+ m2r2 + m3r3 =(9.00 kg)(0 m)? +(6.00 kg)(3.00 m) + (7.00 kg)(5.00 m)2 = 229 kg. m² System B 1 = m_vi_ + m2r2 +m373 (9.00 kg)(5.00 m)+6.00 kg)(4.00 m)2 + (7.00 kg)(0 m)2 = 321 kg. m² b. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque. The torque produced by the force has a magnitude that is equal to the product of the force magnitude and the lever arm: System A T=-Fl=-(424 N)(3.00 m) = -1270 N·m The torque is negative because it produces a clockwise rotation about the axis. System B T=Fl =(424 N)(0 m)=0 N·m c. The final angular velocity w is related to the initial angular velocity 06, the angular acceleration a, and the time t by @= 0, +at The angular acceleration is given by Newton's second law for rotational motion as ar =(ET)/1 where Et is the net torque and I is the moment of inertia. Since there is only one torque acting on each system, it is the net torque, so Et=t. Substituting this expression for a into Equation yields T 0= 0, +at= 0, + 1 In both cases the initial angular velocity is 00 = 0 rad/s, since the systems start from rest. The final angular velocities after 5.00 s are:

T System A 0=0+| -1270 N.m t = (0 rad/s)+1 229 kg.mº (5.00 s) = -27.7 rad's 1 System B 0= 0+1 ON m t=(0 rad/s)+1 321 kg.m ·m? (5.00 s) 15.00 = 0 rad/s

mi 3.00 m m2 Axis 5.00 m 4.00 m m3 System C

d. The moment of inertia for each system is System C 1 = m,??? +m2+2 + +m313 =(9.00 kg)(3.00 m)² +6.00 kg)(0 m)² + (7.00 kg)(4.00m)? 193 kg.mº The torque is negative because it produces a clockwise rotation about the axis. System C T=Fl =(424 N)(0 m)= 0 N·m

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