A 0.290 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. Part A) What is the speed of the potato at the lowest point of its motion?Part B) What is the tension on the string at this point?

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

A) v = 7 m/s

B) T = 8.526N

Explanation:

A) From work energy theorem, we know that;

K1 + U1 + W_other = K2 + U2

Where;

K1 and K2 are initial and final kinetic energy respectively while U1 and U2 are initial and final potential energy respectively.

Now, since the tension in the rope is always perpendicular to the direction of motion, W_other will be zero.

Also, since the potato started from rest, initial kinetic energy will be zero.

Also, since the potato ends at the zero potential level, final potential energy is zero.

Thus, we now have;

U1 = K2

We know that

Potential Energy = mgh

Kinetic energy = (1/2)mv²

Thus, mgh = (1/2)mv²

m will cancel out and we have;

gh = (1/2)v²

Making v the subject;

v² = 2gh

v = √2gh

v = √2 x 9.8 x 2.5

v = √49

v = 7 m/s

B) Since the potato is in uniform circular motion at any point in it's part, it's acceleration is given by;

Angular acceleration; α = v²/R

Now taking the sum of forces at this point, and along the radial direction, if we apply Newton's second law of motion, we'll obtain;

Σ_f = T - W = F_rad

Where;

T is tension in spring,

W is weight = mg

F_rad is normal radial force = mα

Thus, we now have;

Σ_f = T - mg = mα

Now, from earlier, α = v²/R

Thus, T - mg = m(v²/R)

Plugging in the relevant values to get ;

T - (0.29 x 9.8) = 0.29x(7²/2.5)

T - 2.842 = 5.684

T = 5.684 + 2.842

T = 8.526N