Question 298: Work, Energy and Power & Circular Motion

Question

A body is moving along a circular track of radius 100 m with velocity $20 \mathrm{~m} / \mathrm{s}$. Its tangential acceleration is $3 \mathrm{~m} / \mathrm{s}^2$, then its resultant acceleration will be

Accepted Answer

The resultant acceleration of a body moving along a circular track can be determined by combining its tangential acceleration with its centripetal acceleration vectorially.
Tangential Acceleration ($a_t$):
Given as $3 \, \mathrm{m/s}^2$.
Centripetal Acceleration ($a_c$):
This is required to maintain the circular motion and is given by the formula:
$$ a_c = \frac{v^2}{r} $$
where $v = 20 \, \mathrm{m/s}$ is the velocity and $r = 100 \, \mathrm{m}$ is the radius of the circular path.
$$ a_c = \frac{(20)^2}{100} = \frac{400}{100} = 4 \, \mathrm{m/s}^2 $$
Resultant Acceleration ($a_r$):
Since the centripetal and tangential accelerations are perpendicular to each other, the resultant acceleration can be calculated using the Pythagorean theorem:
$$ a_r = \sqrt{a_t^2 + a_c^2} $$
$$ a_r = \sqrt{(3)^2 + (4)^2} $$
$$ a_r = \sqrt{9 + 16} $$
$$ a_r = \sqrt{25} $$
$$ a_r = 5 \, \mathrm{m/s}^2 $$
Thus, the resultant acceleration of the body is $ \boxed{5 \, \mathrm{m/s}^2} $.

Answer

$5 \mathrm{~m} / \mathrm{s}^2$

Answer

$4 \mathrm{~m} / \mathrm{s}^2$

Answer

$2 \mathrm{~m} / \mathrm{s}^2$

Answer

$3 \mathrm{~m} / \mathrm{s}^2$

Work, Energy and Power & Circular Motion

A body is moving along a circular track of radius 100 m with velocity 20 \mathrm{~m} / \mathrm{s}. Its tangential acceleration is 3 \mathrm{~m} / \mathrm{s}^2, then its resultant acceleration will be

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