標題:
phy mech rotation and rolling
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請回答part b問題, thank you http://www.flickr.com/photos/111837627@N03/
最佳解答:
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First assume there is no slipping when the cylinder rolls down the ramp, then we have, for translational motion, 3.2g.sin(30) - Ff = 3.2a ------------- (1) where g is the acceleration due to gravity (= 10 m/s^2) Ff is the (static) frictional force a is the linear acceleration of the centre of mass of the cylinder For rotational motion of the cylinder, Ff.(0.15) = [(3.2 x 0.15^2)/2]a' --------------- (2) where 0.15 m is the radius of the cylinder and the term [(3.2 x 0.15^2)/2] kg.m^2 is the moment of inertia of the cylinder a' is the angular acceleration of the cylinder The condition to satisfy the assumption of no slipping is a = (0.15)a' Hence, (2) becomes: Ff(0.15) = [(3.2 x 0.15^2)/2](a/0.15) i.e. Ff = 1.6a or a = Ff/1.6 Substitute into (1) and solve for Ff give Ff = 5.3 N But max static frictional force that can be provided = (0.3).(3.2g).cos(30) N = 8.3 N Therefore, the static friction is sufficient to prevent no slipping to occur when the cylinder rolls down the ramp. The angular acceleration is thus, using equation (2), equals to 22.1 s^-2 The coefficient of kinetic friction = 5.3/(3.2gcos(30)) = 0.19
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