This preview shows page 1. Sign up to view the full content.
Tm gm v r Tmg gL L mg −= ¡ =+ F H G I K J = 2 4 5 . With m = 0.092 kg, the tension is given by T = 4.5 N. (c) The pendulum is now started (with zero speed) at 90 i θ =° (that is, h i = L ), and we look for an angle such that T = mg . When the ball is moving through a point at angle , then Newton's second law applied to the axis along the rod yields g m v r cos 2 which (since r = L ) implies v 2 = gL (1 – cos θ ) at the position we are looking for. Energy conservation leads to (1 KU KU mgL mv mgL gL gL gL ii += + + − =− + − 0 1 2 1 1 2 1 2 (c o s ) ((c o s ) ) c o s ) θθ where we have divided by mass in the last step. Simplifying, we obtain 1 1 cos 71 3 − §· == ° ¨¸ ©¹ . (d) Since the angle found in (c) is independent of the mass, the result remains the same if the mass of the ball is changed. 109. The connection between angle (measured from vertical) and height
This is the end of the preview. Sign up to access the rest of the document.