Hi, I already have the answers to these questions, but need help on how I can actually get to the answers. Any and all help is greatly appreciated. Thanks!
Problem 1: A 3.0 kg mass is attached to the end of a light rod and made to move clockwise in a vertical circle of radius 2m at a constant speed of 6 m/s. The tension in the rod at the lowest point is most nearly (Answer = 84N).
Problem 2: A block of mass 2 kg is initially at rest when a force F of 4N acts on it at an angle of 45˚ above the horizontal. Consider the horizontal surface to be frictionless. The acceleration of the block due to force F is...? (Answer = 1.4 m/s).
Problem 1: A 3.0 kg mass is attached to the end of a light rod and made to move clockwise in a vertical circle of radius 2m at a constant speed of 6 m/s. The tension in the rod at the lowest point is most nearly (Answer = 84N).
Problem 2: A block of mass 2 kg is initially at rest when a force F of 4N acts on it at an angle of 45˚ above the horizontal. Consider the horizontal surface to be frictionless. The acceleration of the block due to force F is...? (Answer = 1.4 m/s).
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at the lowest point, newton's second law becomes
T - mg = mv^2/r the sum of forces produce the accel term m v^2/r; since the accel acts toward the center of the circle, the centrip force acts up, so we have
T = m(g+v^2/r) = 3kg x (9.8m/s/s + (6m/s)^2/2m) = 83.4N
2. the horizontal force = 4 cos 45 = 2.83N
F = m a so that a = F/m = 2,8N/2kg = 1.4m/s/s
T - mg = mv^2/r the sum of forces produce the accel term m v^2/r; since the accel acts toward the center of the circle, the centrip force acts up, so we have
T = m(g+v^2/r) = 3kg x (9.8m/s/s + (6m/s)^2/2m) = 83.4N
2. the horizontal force = 4 cos 45 = 2.83N
F = m a so that a = F/m = 2,8N/2kg = 1.4m/s/s