find: The diagram above show a 0.340 kg mass tied to a 65.0 cm long string The m

Question

find: The diagram above show a 0.340 kg mass tied to a 65.0 cm long string The mass is released, from rest, from a point where the string (massless and stretchless) makes an angle of 52.0 degree relative to vertical. The mass swings back and forth with negligible air-resistance acting upon it. Taking the mass to have a gravitational potential energy of zero when the string is vertical, what is its gravitational potential energy when it is released? What is its kinetic energy when released? As the mass travels from its release point to its lowest position, how much work Is done on it by the tension in the string? How much work is done on the mass by non-conservative forces as it travels from its release point to its lowest position? Taking the mass to have a gravitational potential energy of zero when the string is vertical, what is its gravitational potential energy when it is at its lowest position? What is the kinetic energy of the mass at the moment when the string is vertical? How fast is the mass traveling at the moment when the string is vertical? What is the magnitude of the mass's acceleration at the moment when the string is vertical? What direction is the mass's acceleration at the moment when the string is vertical? What is the tension in the string at the moment when the string is vertical?

Explanation / Answer

a) Initila height above its lowest point,

h = L*(1 - cos(theta))

so, initial potentail energy, Ui = m*g*h

= m*g*L*(1 - cos(theta))

b) KEi = 0

c) Work done by tension = T*d*cos(90)

= 0 (angle between tension and dispalcement is always 90 degrees)

d) W(non-conservative force) = 0

e) U_lowest = -m*g*L*(1 - cos(theta))

f) KE_lowest = m*g*L*(1 - cos(theta))

g) v = sqrt(2*g*L*(1 - cos(theta) )

h) a_rad = v^2/L

= 2*g*L*(1 - cos(theta))/L

= 2*g*(1 -cos(theta))

i) upward

j) T = m*g + m*a_rad

= m*g + m*2*g*(1 -cos(theta))

= m*g*(1 + 2 - cos(theta))

= m*g*(3 - cos(theta))

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