A simple pendulum consists of an object suspended by a string. The object is ass

Question

A simple pendulum consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air resistance the system oscillates by swinging back and forth in a vertical plane. If the string is 2.05 m long and makes an initial angle of 35.0 degrees with the vertical, calculate the speed of the particle at the following positions.

(a) at the lowest point in its trajectory

(b) when the angle is 15.0 degrees

Explanation / Answer

If you look at the picture in the "sources" section, there is a triangle with Angle A, side L-h, and hypotenuse L. Therefore: cos(A) = (L-h)/L h = L - L*cos(A) = L*(1-cos(A)) Now, we can find the potential energy at any height: PE = mgh = m*g*L*(1-cos(A)) Kinetic Energy = KE = (1/2) * m * v^2 At the initial angle of A = 35°, the velocity is zero, so the total energy: KE_35 + PE_35 = 0 + m*g*L*(1-cos(35°)) = m*9.81*2.05*0.18 = 3.62 * m (has units of J/kg) We use this value as our total energy for any other point, since there is no friction or air resistance, so the total energy remains constant. At bottom of path: PE = 0, KE = (1/2) * m * (v_bottom)^2 KE_bottom = PE_35 ==> (1/2) * m * (v_bottom)^2 = 3.62 * m Solve for v_bottom: v_bottom = sqrt(2*3.62 ) = 2.69 m/s At angle 15°: PE_15 = m*g*L*(1-cos(15°)) = m * 9.81 * 2.10 * (.03407) = .70196 * m KE_15 = (1/2) * m * (v_15)^2 PE_15 + KE_15 = PE_35 (1/2) * m * (v_15)^2 = (3.62 * m) - (.70196 * m) The m's cancel. Solve for v_15: (v_15) = sqrt[ 2* (3.62 - .70196) ] = 2.4158 m/s

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