Question 14: Work, Energy and Power & Circular Motion

Question

A ball is moving in a circular path of radius $$5 \mathrm{~m}$$. If tangential acceleration at any instant is $$10 \mathrm{~ms}^{-2}$$ and the net acceleration makes an angle of $$30^{\circ}$$ with the centripetal acceleration, then, the instantaneous speed is

Accepted Answer

To solve this problem, we'll use the relationship between tangential acceleration, centripetal acceleration, and the net acceleration of an object moving in a circular path.
The given details are:
Radius of the circular path, $$r = 5 \, 	ext{m}$$
Tangential acceleration, $$a_t = 10 \, 	ext{m/s}^2$$
Angle between net acceleration and centripetal acceleration, $$	heta = 30^\circ$$
First, let's denote the centripetal acceleration as $$a_c$$ and the net acceleration as $$a_{net}$$.
We know that the net acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. Since the angle between the net acceleration and the centripetal acceleration is given as $$30^\circ$$, we can use trigonometric relationships to solve for the centripetal acceleration.
From the definition of net acceleration, we can establish:
$$a_{	ext{net}} = \sqrt{a_c^2 + a_t^2}$$
Given that:
$$\cos 	heta = \frac{a_c}{a_{	ext{net}}}$$
Substituting the values we get:
$$\cos 30^\circ = \frac{a_c}{\sqrt{a_c^2 + a_t^2}}$$
Knowing $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$, we substitute it into the equation:
$$\frac{\sqrt{3}}{2} = \frac{a_c}{\sqrt{a_c^2 + (10 \, 	ext{m/s}^2)^2}}$$
Rearranging and squaring both sides, we get:
$$(\frac{\sqrt{3}}{2})^2 = \frac{a_c^2}{a_c^2 + 100}$$
This simplifies to:
$$\frac{3}{4} = \frac{a_c^2}{a_c^2 + 100}$$
Cross-multiplying gives:
$$3(a_c^2 + 100) = 4a_c^2$$
$$3a_c^2 + 300 = 4a_c^2$$
$$300 = a_c^2$$
$$a_c = \sqrt{300} = 10\sqrt{3} \, 	ext{m/s}^2$$
Now, identifying that centripetal acceleration is given by:
$$a_c = \frac{v^2}{r}$$
Substituting values, we have:
$$10\sqrt{3} = \frac{v^2}{5}$$
Solving for $$v$$ gives:
$$10\sqrt{3} 	imes 5 = v^2$$
$$v^2 = 50\sqrt{3}$$
$$v = \sqrt{50\sqrt{3}}$$
Converting and simplifying:
$$v = (\sqrt{50}) (\sqrt[4]{3})$$
$$v \approx 7.07  \cdot 1.32 = 9.3 $$
Thus, the instantaneous speed is $$9.3 \, 	ext{m/s}$$.
The correct answer is:
Option D: $$9.3 \mathrm{~ms}^{-1}$$

Answer

$$5.4 \mathrm{~ms}^{-1}$$

Answer

$$50 \sqrt{3} \mathrm{~ms}^{-1}$$

Answer

$$6.6 \mathrm{~ms}^{-1}$$

Answer

$$9.3 \mathrm{~ms}^{-1}$$

Work, Energy and Power & Circular Motion

A ball is moving in a circular path of radius $5 \mathrm{~m}. If tangential acceleration at any instant is 10 \mathrm{~ms}^{-2} and the net acceleration makes an angle of 30^{\circ}$ with the centripetal acceleration, then, the instantaneous speed is

Choose the correct answer: