Expert Answer to "A 0.50 kg ball that is tied to..."

High School
Physics
Force, Motion and Energy
Acceleration Due to Gravity

Harlen Pritchard

Answered question

2020-11-05

A 0.50 kg ball that is tied to the end of a 1.1 m light cord is revolved in a horizontal planewith the cord making a $30^{\circ}$ angle, with the vertical.

a) Determine the ball speed.
b) If, instead, the ball is revolved so that its speed is 4.0 m/s,what angle does the cord make with the vertical?
c) If the cord can withstand a maximum tension of 9.1 N, what is the highest speed at which the ballcan move?

Answer & Explanation

Aubree Mcintyre

Skilled2020-11-06Added 73 answers

a) $\frac{m v^{2}}{r} = T \sin (θ)$
$m g = T \cos (θ)$
$\tan (θ) = \frac{v^{2}}{r g}$
$v^{2} = r g \tan (θ)$
$r = h \tan (θ)$
$v^{2} = h {g (\tan (θ))}^{2}$
$v^{2} = 1.1 \times 9.8 \times {(\tan (30))}^{2}$ $= 3.5933$
$v = 1.8956 \frac{m}{s}$
b): $v^{2} = h {g (\tan (θ))}^{2}$
${(\tan (θ))}^{2} = \frac{4^{2}}{1.1 \times 9.8}$
$θ = 50.62$ degrees
c): $\frac{m v^{2}}{r} = T \sin (θ)$
$v^{2} = \frac{r T \sin (θ} {m}}{}$ $= \frac{T h \tan (θ) \sin (θ)}{m}$ $= \frac{9.1 \times 1.1 \times \tan (30) \sin (30)}{0.5 k g}$
$v = 2.404 \frac{m}{s}$

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