A pendulum consists of a massless string with length L with one end attached to

Question

A pendulum consists of a massless string with length L with one end attached to a fixed point and a mass m attached to the free end (see figure). Initially the line is held fully stretched forming an angle of 45 with the vertical and then let go. Give answers in terms of m, L, and g. For parts c) and d) you’ll also use in your answers. [a] Find the speed of the mass when the line is vertical. Check units. [b] Find the tension of the string when the line is vertical. Check units. [c] Find the speed of the mass when the line is at an angle with respect to the vertical (where < 45 ). [d] Find the work done by gravity to move the mass from the initial position to an angle with respect to the vertical (where < 45 ). Check units. [e] Find the work done by tension to move the mass from the initial position to an angle with respect to the vertical (where < 45 ). Check units.

Explanation / Answer

initial vertical height = h1 = L*(1-cos45) = L(1-1/sqrt2) = 0.293 L

initial potential energy PE1 = m*g*h1 = 0.293 *m*g*L

in vertical position KE = 0.5*m*v^2

from energy conservation

KE in vertical = PE at45

0.5*m*v^2 = 0.293*m*g*L

v = sqrt(0.586*g*L)      <,---------answer

(b)

T = mg + mv^2/L

T = mg + m*0.586*g* = 1.586 mg <,-----answer

(c)

vertiacl height at theta = h2 = L*(1 - costheta)

PE = m*g*h2

kinetic energy KE = 0.5*m*v^2

initial potential energy = PE2 = m*g*h1

from energy conservation

m*g*h1 = m*g*h2 + 0.5*m*v^2

m*g*L*(1-cos45 ) = m*g*L*(1-costheta) + 0.5*m*v^2

2*g*L*(costheta - cos45 ) = v^2

v = sqrt(2*g*L*(costheta - cos45 )) <<<--------------answer

(d)

Wg = m*g*h(h2-h1)

Wg = m*g*L*(costheta - cos45)

(e)

T is penependicular to the the displacement

Wt = T*s*cos90 = 0 <<<-----------answer

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