Question

One end of a cord is fixed and a small 0.500-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m as shown in the Figure. When θ = 20.0°, the speed of the object is 8.00 m/s. At this instant, find (a) the tension in the string, (b) the tangential and radial components of acceleration, and (c) the total acceleration.

Answer 1

Answer on Question #73984, Physics / Mechanics | Relativity

Question. One end of a cord is fixed and a small $0.500 \, kg$ object is attached to the other end, where it swings in a section of a vertical circle of radius $2.00 \, m$ as shown in the Figure. When $\theta = 20.0{}^\circ$ , the speed of the object is $8.00 \, m/s$ . At this instant, find

(a) the tension in the string;

(b) the tangential and radial components of acceleration, and

(c) the total acceleration.

Solution.

Given. $m = 0.500kg$ ; $r = 2.00m$ ; $\theta = 20.0{}^{\circ}$ ; $v = 8.00m / s$ .

Find. $T, a_{n}, a_{\tau}, a - ?$

Solution.

(a) the tension in the string

According to the Second Newton's law

$$\sum \vec {F} = m \vec {a}.$$

We have

$$T - m g \cos \theta = m \frac {v ^ {2}}{r} \rightarrow T = m \frac {v ^ {2}}{r} + m g \cos \theta = 0. 5 \cdot \frac {8 ^ {2}}{2} + 0. 5 \cdot 9. 8 1 \cdot \cos 2 0 {}^ {\circ} = 2 0. 6 0 9 N$$

(b) the tangential and radial components of acceleration

the tangential component

$$mg \sin \theta = ma_{\tau} \rightarrow a_{\tau} = g \sin \theta = 9.81 \cdot \sin 20{}^{\circ} = 3.35 \, \mathrm{m/s^2}.$$

the radial component

$$a_{n} = \frac{v^{2}}{r} = \frac{8^{2}}{2} = 32 \, \mathrm{m/s^2}.$$

(c) the total acceleration

$$a = \sqrt{a_{n}^{2} + a_{\tau}^{2}} = \sqrt{32^{2} + 3.35^{2}} = 32.2 \, \mathrm{m/s^2}.$$

Answer. $T = 20.609 \, \mathrm{N}; a_{\tau} = 3.35 \, \mathrm{m/s^2}; a_{n} = 32 \, \mathrm{m/s^2}; a = 32.2 \, \mathrm{m/s^2}$.

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