12_P67InstructorSolution

12_P67InstructorSolution - 12.67. Model: The hollow...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
12.67. Model: The hollow cylinder is a rigid rotating body. Visualize: We placed the origin of the coordinate system on the ground. Solve: (a) Newton’s second law for the block is GB B () , y FT m a −+ = where T is the tension in the string, B Fm g = is the gravitational force on the block, and y a is the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow cylinder is 2 y y aI a TR I I T R R τα =− = = ⇒ =− where we used the acceleration constraint . y aR α = With this expression for T , Newton’s second law for the block becomes B BB 22 B (/ ) yy Ia m g mg ma a R mI R −− = = + The moment of inertia of a hollow cylinder is 2 C , Im R = so the equation for y a simplifies to 2 2 B BC (3.0 kg)(9.8 m/s ) 5.88 m/s 3.0 kg 2.0 kg y a mm == = ++ The speed of the block just before it hits the ground can now be found using kinematics:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online