Question

In a carnival ride the passengers travel in a circle of radius $5.0\;{\text{m}}$, making one complete circle in $4.0\;{\text{s}}$. What is the acceleration?
(A) $24.6\;{\text{m/}}{{\text{s}}^{\text{2}}}$
(B) $12.3\;{\text{m/}}{{\text{s}}^{\text{2}}}$
(C) $6.15\;{\text{m/}}{{\text{s}}^{\text{2}}}$
(D) $49.2\;{\text{m/}}{{\text{s}}^{\text{2}}}$

Answer 1

Hint:- A direction of the centripetal force is towards the center of the circle. The distance travelled will be the circumference of the circle. And the expression for the centripetal acceleration can be determined from the centripetal force.

Complete step by step answer:
Given the radius of the circle is $r = 5.0\;{\text{m}}$
The expression for the circumference of the circle is given as,
$C = 2\pi r$
Where, $r$ is the radius of the circle.
Substituting the values in the above expression,
$\ C = 2 \times 3.14 \times 5\;{\text{m}} \\\ {\text{ = 31}}{\text{.4}}\;{\text{m}} \\\ \$
Therefore the distance travelled by the passenger is ${\text{31}}{\text{.4}}\;{\text{m}}$.
Given the time taken for the passenger making one complete rotation of circle is $t = 4.0\;{\text{s}}$
The velocity is the distance traveled by the passenger in time $t$ .
The expression for the velocity is given as,
$v = \dfrac{d}{t}$
Here, $d$ is the circumference of the circle passenger traveled.
Substituting the values in the above expression,
$\ v = \dfrac{{{\text{31}}{\text{.4}}\;{\text{m}}}}{{4\;{\text{s}}}} \\\ = 7.85\;{\text{m/}}{{\text{s}}^{\text{2}}} \\\ \$
Therefore the velocity of the passenger in the circular path is $7.85\;{\text{m/}}{{\text{s}}^{\text{2}}}$.
The centripetal force required by passenger to maintain the circular path is given as,
${f_C} = \dfrac{{m{v^2}}}{r}$
Where, $m$ is the mass of the passenger, $r$ is the radius of the circle.
According to Newton’s second law of motion, the force is defined as the mass times the acceleration of the body. Thus comparing the above expression with Newton’s second law of motion, the expression for the centripetal acceleration is given as,
$a = \dfrac{{{v^2}}}{r}$
Substituting the values in above expression,

a = \dfrac{{{{\left( {7.85\;{\text{m/}}{{\text{s}}}} \right)}^2}}}{{5\;{\text{m}}}} \\\ = 12.3\;{\text{m/}}{{\text{s}}^{\text{2}}} \\\ \ $$ Therefore the acceleration of the passenger is $$12.3\;{\text{m/}}{{\text{s}}^{\text{2}}}$$. **The answer is option B.** **Note:** We have to note that the radius of the circular path is a factor of centripetal acceleration. The acceleration will be maximum for smaller circles and minimum for long circles.