Answer
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Hint: It is required to initially know the types of forces acting on the wheel at any point on wheel. Then substitute the forces acting towards the centre of the wheel by the forces acting outward. This is done so as centripetal force acts towards the centre of the wheel giving a net inward push. Applying values in the equation will lead to the answer.
Formulas used:
${F_C} = \dfrac{{m{v^2}}}{r}$
$\Rightarrow F = ma$
$\Rightarrow a = \dfrac{{{v^2}}}{r}$
Complete step by step answer:
Centripetal force is the force that makes the object move in circular motion as it acts towards the centre and doesn’t allow the object to leave the trajectory. This force produces a tangential acceleration called centripetal acceleration. It can be defined as the rate of change of tangential velocity. Centripetal force is given by:
${\text{Centripetal Force}} = {F_C} = \dfrac{{m{v^2}}}{r}$
Where, $m = {\text{mass}}$, $v = {\text{velocity}}$ and $r = {\text{radius}}$
Now as we know, according to Newton’s second law of motion:
$F = ma$
Where, $m = {\text{mass }}$ and $a = {\text{acceleration}}$.
Comparing both equations and putting values of force equal we obtain,
$F = ma = \dfrac{{m{v^2}}}{r}$
Obtaining equation of acceleration as,
$a = \dfrac{{{v^2}}}{r}$ ……..(i)
As the answer is to be in terms of $m\,{s^{ - 1}}$ hence we need to convert the radius of the wheel into units of meter $\left( m \right)$ .
Given terms,
$\text{Diameter} = 44\,cm = 0.44\,m$
$\text{Radius} = \dfrac{\text{Diameter}}{2} = \dfrac{{0.44}}{2} = 0.22\,m$
$\text{Velocity} = v = 2.5\,m\,{s^{ - 1}}$
Putting values of radius and velocity in equation of acceleration (i),
$a = \dfrac{{2.5 \times 2.5}}{{0.22}} \\
\therefore a= 28.41\,m\,{s^{ - 2}}$
Centripetal acceleration on the rim of the wagon is $28.41\,m\,{s^{ - 2}}$.
Hence, the correct answer is option C.
Note: It is always mandatory to convert all units into the same dimensions. Non conversion can yield a mis-match of dimensions and give an error in result. Circular motion is not only about centripetal forces but is resultant of several forces such as centrifugal, weight, friction etc. In the mentioned question resistant forces are omitted.
Formulas used:
${F_C} = \dfrac{{m{v^2}}}{r}$
$\Rightarrow F = ma$
$\Rightarrow a = \dfrac{{{v^2}}}{r}$
Complete step by step answer:
Centripetal force is the force that makes the object move in circular motion as it acts towards the centre and doesn’t allow the object to leave the trajectory. This force produces a tangential acceleration called centripetal acceleration. It can be defined as the rate of change of tangential velocity. Centripetal force is given by:
${\text{Centripetal Force}} = {F_C} = \dfrac{{m{v^2}}}{r}$
Where, $m = {\text{mass}}$, $v = {\text{velocity}}$ and $r = {\text{radius}}$
Now as we know, according to Newton’s second law of motion:
$F = ma$
Where, $m = {\text{mass }}$ and $a = {\text{acceleration}}$.
Comparing both equations and putting values of force equal we obtain,
$F = ma = \dfrac{{m{v^2}}}{r}$
Obtaining equation of acceleration as,
$a = \dfrac{{{v^2}}}{r}$ ……..(i)
As the answer is to be in terms of $m\,{s^{ - 1}}$ hence we need to convert the radius of the wheel into units of meter $\left( m \right)$ .
Given terms,
$\text{Diameter} = 44\,cm = 0.44\,m$
$\text{Radius} = \dfrac{\text{Diameter}}{2} = \dfrac{{0.44}}{2} = 0.22\,m$
$\text{Velocity} = v = 2.5\,m\,{s^{ - 1}}$
Putting values of radius and velocity in equation of acceleration (i),
$a = \dfrac{{2.5 \times 2.5}}{{0.22}} \\
\therefore a= 28.41\,m\,{s^{ - 2}}$
Centripetal acceleration on the rim of the wagon is $28.41\,m\,{s^{ - 2}}$.
Hence, the correct answer is option C.
Note: It is always mandatory to convert all units into the same dimensions. Non conversion can yield a mis-match of dimensions and give an error in result. Circular motion is not only about centripetal forces but is resultant of several forces such as centrifugal, weight, friction etc. In the mentioned question resistant forces are omitted.
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