Posted by Gabe on Friday, January 22, 2010 at 3:17am.
Let T be the string tension. That and the angular acceleration alpha are the unknowns. Separate equations for the acceleration of the block and the angular acceleration of the cylinder will allow you to solve for both variables. The moment of inertia of the cylinder in (1/2) M r^2
m g - T = m a
T*r = torque = I*alpha = (1/2) m r^2 * alpha
T = (1/2) m*r*alpha
m g = ma + (1/2) m r * alpha
= ma + (1/2) m a
(since alpha = a/r)
a = (2/3) g
alpha = (2/3)(g/r)
Related Questions
physcs - A string is wrapped around a uniform solid cylinder of radius 3.10 cm, ...
physics - A long string is wrapped around a 6.2cm diameter cylinder, initially ...
Physics - A massless string is wrapped around a cylinder of mass 0.400 kg and ...
Physics - Consider a stainless steel annular disk with an outer radius 66 mm and...
Physics - A 2.9-kg 12-cm-radius cylinder, initially at rest, is free to rotate ...
Physics - I need help with this question please. A solid cylinder of mass m and ...
Physics - Consider a stainless steel annular disk with an outer radius 68mm and ...
Physics - Consider a stainless steel annular disk with an outer radius 68mm and ...
Physics - I need help with this question please. A solid cylinder of mass m and...
Applied physics - a string is wrapped around a cylinder with radius of 20cm. If ...
For Further Reading