Question

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

• A. \( 5.4 \, \text{m/s} \)
• B. \( 50\sqrt{3} \, \text{m/s} \)
• C. \( 6.6 \, \text{m/s} \)
• D. \( 9.3 \, \text{m/s} \)

Hint:

When dealing with circular motion problems involving tangential and centripetal acceleration, remember that the net acceleration is the vector sum of these two accelerations, and you can use trigonometric relationships to solve for unknown quantities.

Answer 1

The Correct Option is D

The net acceleration \( a_{\text{net}} \) can be expressed as the vector sum of the tangential acceleration \( a_t \) and the centripetal acceleration \( a_c \). The centripetal acceleration is given by: \[ a_c = \frac{v^2}{r} \] Where: 
- \( v \) is the instantaneous speed of the ball, 
- \( r = 5 \, \text{m} \) is the radius. Given: 
- The tangential acceleration \( a_t = 10 \, \text{m/s}^2 \), 
- The angle between the net acceleration and the centripetal acceleration is \( 30^\circ \). Using the relation for the net acceleration: \[ a_{\text{net}}^2 = a_c^2 + a_t^2 \] Also, the net acceleration is related to the angle \( \theta \) between the tangential and centripetal accelerations: \[ a_{\text{net}} = \sqrt{a_t^2 + a_c^2} \] Since \( \theta = 30^\circ \), the tangent of the angle gives: \[ \tan(30^\circ) = \frac{a_t}{a_c} \] This gives: \[ \frac{10}{a_c} = \tan(30^\circ) = \frac{1}{\sqrt{3}} \] Solving for \( a_c \): \[ a_c = 10\sqrt{3} \, \text{m/s}^2 \] Now substitute the value of \( a_c \) into the centripetal acceleration formula: \[ a_c = \frac{v^2}{r} \Rightarrow 10\sqrt{3} = \frac{v^2}{5} \] Solving for \( v \): \[ v^2 = 50\sqrt{3} \] \[ v = \sqrt{50\sqrt{3}} \approx 9.3 \, \text{m/s} \] 
Thus, the instantaneous speed is \( 9.3 \, \text{m/s} \).