{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

12_P67InstructorSolution

# 12_P67InstructorSolution - 12.67 Model The hollow cylinder...

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

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 G B B ( ) , y F T m a + = where T is the tension in the string, G B B ( ) F m 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 a Ia TR I I T R R τ α = − = = = − where we used the acceleration constraint . y a R α = With this expression for T , Newton’s second law for the block becomes B B B 2 2 B ( / ) y y Ia m g m g m a a R m I R = = + The moment of inertia of a hollow cylinder is 2 C , I m R = so the equation for y a simplifies to 2 2 B B C (3.0 kg)(9.8 m/s ) 5.88 m/s 3.0 kg 2.0 kg y m g a m m = = = − + + The speed of the block just before it hits the ground can now be found using kinematics:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online