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 friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.00 m long and makes an initial angle of 26.0° 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°

Explanation / Answer

Here are our given values:

Length of pendulum: L = 2 m
Initial angle: A_initial = 26°

It doesn't mention it in the problem, but let's assume the particle has mass "m".

Find: v_bottom, v_15°

We can find the height of the particle, at a given angle. 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 = 26°, the velocity is zero, so the total energy:

KE_26 + PE_26 = 0 + m*g*L*(1-cos(26°)) = m*9.81*2*(1-cos(26°)) = 1.986 * m (1.986 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_26 ==> (1/2) * m * (v_bottom)^2 = 1.986 * m

Solve for v_bottom: v_bottom = sqrt(2*1.986) = 1.9929 m/s

At angle 15°:

PE_15 = m*g*L*(1-cos(15°)) = m * 9.81 * 2 * (1-cos(15°)) = .668 * m

KE_15 = (1/2) * m * (v_15)^2

PE_15 + KE_15 = PE_26

(1/2) * m * (v_15)^2 = (1.986 * m ) - (.668 * m)

The m's cancel. Solve for v_15:

(v_15) = sqrt[ 2* (1.986 - .668 )] = 1.62 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.