Question

I’m slightly off for part A and I dont understand what part B is asking me. Beyond confused,

REMARKS The moment of inertia is smaller in part (B) because in this configuration the 0.20-kg spheres are essentially located on the axis of rotation. QUESTION If one of the rods is lengthened, which one would cause the larger change in the moment of inertia? the rod connecting balls two and four the rod connecting the bars one and three PRACTICE IT Use the worked example above to help you solve this problem. In an effort to be the star of the half-time show, a majorette twirls an unusual baton made up of four spheres fastened to the end of very light rods (see figure). Each rod is 1.30 m long. (a) Find the moment of inertia of the baton about an axis perpendicular to the page and passing through the point where the rods cross. 04632X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. kg tm (b) The majorette tries spinning her strange baton about the axis 00, as shown in the figure Calculate the moment of inertia of the baton about this axis kg sim ME in both cases, vector T, and vector T, exert torgues on the pultey.O PROBLEM Two blocks with masses m1° 5.00 kg and m2 7.00 kg are attached by a string as in Figure (a), over a pulley with mass M 2.00 kg. The pulley, which turns on a frictionless axle, is a hollow cylinder with radius 0.0500 m over which the string moves without slipping. The horizontal surface has coefficient of kinetic friction 0.350. Find the speed of the system when the block of mass m2 has dropped 2.00 m STRATEGY This problem can be solved with the extension of the work-energy theorem. If the block of mass m2 falls from height h to o, then the block of mass m1 moves the same distance, Δ.t h. Apply the work-energy theorem, solve for v, and substitute, Kinetic friction is the sole nonconservative force

Answer 1

A)

I = (0.2 x 0.65^2) + (0.2 x 0.65^2) + (2 x 0.3 x 0.65^2)

I = 0.4225 kg m^2

B)

I = 0.3 x 0.65^2 + 0.3 x 0.65^2

I = 0.254 kgm^2

Comment in case any doubt please rate my answer...