1. The problem statement, all variables and given/known data 1. A wheel mounted on an axis that is not frictionless is initially at rest. A constant external torque of 50 N*m is applied to the wheel for 20s, giving the wheel an angular velocity of 600 rev/min. The external torque is then removed, and the wheel comes to rest 120s later. Find the moment of inertia of the wheel, and the frictional torque, which is assumed to be constant. 2.A 2kg block and an 8kg block are both attached to an ideal spring for which k=200 N/m and both are initially at rest on a horizontal frictionless surface. In an initial experiment a 0.1kg ball of clay is thrown at the 2kg block. The clay is moving horizontally with speed v when it hits and sticks to the block. The 8kg block is held still by a removable stop. As a result, the spring compresses a maximum distance of 0.4m. In a second experiment, an identical ball of clay is thrown at another identical 2kg block, but this time the stop is removed so that the 8kg block is free to move. e. State the principle(s) that can be used to calculate the velocity of the 8kg block at the instant that the spring regains its original length. Write the appropriate equation(s) and show the numerical substitutions, but do not solve for velocity. 2. Relevant equations [tex]\tau[/tex]=II*[tex]\alpha[/tex] Ei=Ef Pi=Pf 3. The attempt at a solution 1. I found out the angular acceleration by using change in angular speed over change in time which is 3.14 rad/s^2 Then I set up the system of equation as [tex]\tau[/tex]friction= I*3.14 and 50-[tex]\tau[/tex]friction=5*3.14 Then I found out frictional torque is 25 N*m and I=7.96 kg*m^2 However, my result was wrong so can you guys help me spot out where did I do wrong and how to fix it ? 2.So I use conservation of energy and conservation of momentum. For momentum I have m(clay)*v(clay) + m(blocks)*v(blocks)= (m(clay)+2)+v(f,1) + 8*v(f,2) I substitute numbers in as: 0.1*8.2= 2.1*v(f,1)+8*v(f,2) For conservation of energy I have (1/2)*0.1*8.2^2=(1/2)*2.1*v(f,1)^2+(1/2)*8*v(f,2)^2 I get 336.2=1.05*v(f,1)^2+4*v(f,2)^2 I set a system of equations containing two equations I just got and get v(f,1)= -15.03 and v(f,2)= (8.2-2.1*-15.03)/8 Another wrong result. Anyone can spot out and help me fix the mistake ?? Thank you.