Question #116390
2. A ball on the end of a string is revolved at uniform rate in vertical circle of radius 72.0
cm. If its speed is 4.00 m/s and mass of 0.3 kg. Calculate the tension
in the string when ball is:
a) At top of its path
b) At bottom of its path
1
Expert's answer
2020-05-19T10:40:00-0400


We write the equation of motion of the ball in a circle

N+mg=ma\vec{N}+m\vec{g}=m \vec{a}

Where

a=v2ra=\frac{v^2}{r}

a) We write the equation of motion for the upper position

N+mg=mv2rN+mg=\frac{mv^2}{r}

Then the tension in the string is

N=mv2rmg=0.3420.720.39.81=3.724 NN=\frac{mv^2}{r}-mg=\frac{0.3 \cdot 4^2}{0.72}-0.3 \cdot 9.81=3.724\space{N}

b) We write the equation of motion for the lower position

N+mg=mv2r-N+mg=-\frac{mv^2}{r}

Then the tension in the string is

N=mv2r+mg=0.3420.72+0.39.81=9.61 NN=\frac{mv^2}{r}+mg=\frac{0.3 \cdot 4^2}{0.72}+0.3 \cdot 9.81=9.61\space{N}


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