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John O' Meara
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A light rigid rod 1.0m long has a small block of mass.05kg attached at one end. The other end is pivoted, and the rod rotates in a vertical circle. At a certain instant, the rod is 36.9 degrees above the horizontal, and the tangential speed of the block is 4m/s. (a) What are the horizontal and vertical components of the velocity of the block? (b) What is the moment of inertia of the block? (c) What is the radial acceleration of the block? (d) What is the tangential acceleration of the block? (e) What is the tension or compression in the rod?
Answers: (a) 3.2m/s and 2.4m/s.
(b) L=Iw => I=m*v*r/w=.05kg.m^2.
(c) a_radial=v^2/r=16 rad/s^2.
(d) a_tangential = r*(alpha), where alpha = angular acceleration. This is where my problem is.
(e) Fr=m*a_radial=.8N outwards. It is compressed by m*g*sin(36.9)=.294N, therefore T=.506N.
Where a_ is acceleration, Fr is radial force, w=angular velocity, L is angular momentum, mass=m, velocity = v. Many thanks.
Answers: (a) 3.2m/s and 2.4m/s.
(b) L=Iw => I=m*v*r/w=.05kg.m^2.
(c) a_radial=v^2/r=16 rad/s^2.
(d) a_tangential = r*(alpha), where alpha = angular acceleration. This is where my problem is.
(e) Fr=m*a_radial=.8N outwards. It is compressed by m*g*sin(36.9)=.294N, therefore T=.506N.
Where a_ is acceleration, Fr is radial force, w=angular velocity, L is angular momentum, mass=m, velocity = v. Many thanks.