Answer:
A) The speed of the potato at the lowest point of its motion is 7.004 m/s
B) The tension on the string at this point is 8.5347 N
Explanation:
Here we have that the height from which the potato is allowed to swing is 2.5 m
Therefore we have ω₂² = ω₁² + 2α(θ₂ - θ₁)
Where:
ω₂ = Final angular velocity
ω₁ = Initial angular velocity = 0 rad/s
α = Angular acceleration
θ₂ = Final angle position
θ₁ = Initial angle position
However, we have potential energy of the potato
= Mass m×Gravity g× Height h
= 0.29×9.81×2.5 = 7.1125 J
At he bottom of the swing, the potential energy will convert to kinetic energy as follows
K.E. = P.E. = 7.1125 J
1/2·m·v² = 7.1125 J
Therefore,
v² = 7.1125 J/(1/2×m) = 7.1125 J/(1/2×0.290) = 49.05
∴ v = √49.05 = 7.004 m/s
B) Here we have the tension given by
Tension T in the string = weight of potato + Radial force of motion
Weight of potato = mass of potato × gravity
Radial force of motion of potato = mass of potato × α,
where α = Angular acceleration = v²/r and r = length of the string
∴ Tension T in the string = m×g + m×v²/r = 0.290×(9.81 + 7.004²/2.5)
T = 8.5347 N
