Question

A toy car starts from rest along the circular track of radius ${{R}}$ on a horizontal board with a tangential acceleration ${{{a}}_{{0}}}$. At the same instant, the board starts from rest and accelerates upward with an acceleration ${{{a}}_{{1}}}$. The coefficient of friction between board and car is ${{\mu }}$. The speed of the car relative to the board at which it will skid is.
A. {\left[ {\left\\{ {{{{\mu }}^{{2}}}{{\left( {{{g + }}{{{a}}_{{1}}}} \right)}^{{2}}}{{ - }}{{{a}}_{{0}}}^{{2}}} \right\\}{{{R}}^{{2}}}} \right]^{\dfrac{{{1}}}{{{4}}}}}
B. {\left[ {{{\left\\{ {{{{\mu }}^{{2}}}{{\left( {{{g + }}{{{a}}_{{1}}}} \right)}^{{2}}}{{ - }}{{{a}}_{{0}}}^{{2}}} \right\\}}^{\dfrac{{{1}}}{{{4}}}}}{{{R}}^2}} \right]^{\dfrac{1}{4}}}
C. {\left[ {\left\\{ {{{\mu }}\left( {{{g + }}{{{a}}_{{1}}}} \right){{ - }}{{{a}}_{{0}}}} \right\\}{{{R}}^{{2}}}} \right]^{\dfrac{{{1}}}{{{4}}}}}
D. {\left[ {\left\\{ {{{{\mu }}^{{2}}}\left( {{{{a}}_{{1}}}^{{2}}{{ - }}{{{a}}_{{0}}}^{{2}}} \right)} \right\\}{{{R}}^{{2}}}} \right]^{\dfrac{{{1}}}{{{4}}}}}

Answer 1

This could be simply solved by simply equating the forces on both sides. Also, we should never forget to assume forces due to normal and tangential acceleration separately.

Formula used: Here, we will use the basic formula NLM-2:
${{F = ma}}$
Here, ${{F}}$ is the force exerted on the car
${{m}}$ is the mass of the car
${{a}}$ is acceleration of the car
And
${{{a}}_{{1}}}{{ = }}\dfrac{{{{{v}}^{{2}}}}}{{{r}}}$
Here, ${{{a}}_{{1}}}$ is the normal acceleration
${{v}}$ is the velocity
And ${{r}}$ is the radius

Complete step by step answer:
We already know that the car is experiencing normal as well as tangential acceleration,
So, let us assume the mass of the car to be ${{m}}$.
First, we will calculate the resultant acceleration of the car,
We have tangential acceleration ${{{a}}_{{0}}}$ and normal acceleration ${{{a}}_{{1}}}$
${{{a}}_{{{res}}}}{{ = }}\sqrt {{{{a}}_{{1}}}^{{2}}{{ + }}{{{a}}_{{o}}}^{{2}}}$
$\Rightarrow {{{a}}_{{{res}}}}{{ = }}\sqrt {{{\left( {\dfrac{{{{{v}}^{{2}}}}}{{{r}}}} \right)}^{{2}}}{{ + }}{{{a}}_{{o}}}^{{2}}}$
The coefficient of friction between board and car is ${{\mu }}$.
To stop skidding:
${{{F}}_{{{net}}}}{{ = \mu mg}}$
Also, we know:
${{F = ma}}$
$\Rightarrow {{m \times }}\sqrt {{{\left( {\dfrac{{{{{v}}^{{2}}}}}{{{r}}}} \right)}^{{2}}}{{ + }}{{{a}}_{{o}}}^{{2}}} {{ = \mu mg}}$
On cancelling mass on both sides,
$\dfrac{{{{{v}}^{{4}}}{{ + }}{{{a}}_{{0}}}^{{2}}{{{R}}^{{2}}}}}{{{{{R}}^{{2}}}}}{{ = }}{{{\mu }}^{{2}}}{{{g}}^{{2}}}$
On solving, the above equations,
${{v = [(}}{{{\mu }}^{{2}}}{{{g}}^{{2}}}{{ - }}{{{a}}_{{0}}}^{{2}}{{)}}{{{R}}^{{2}}}{{{]}}^{\dfrac{{{1}}}{{{4}}}}}$
Now, we rearrange:
v = {\left[ {{{\left\\{ {{{{\mu }}^{{2}}}{{\left( {{{g + }}{{{a}}_{{1}}}} \right)}^{{2}}}{{ - }}{{{a}}_{{0}}}^{{2}}} \right\\}}^{\dfrac{{{1}}}{{{4}}}}}{{{R}}^2}} \right]^{\dfrac{1}{4}}}
Then we need to match the correct option.
Hence, the correct option is B.

Note: It should always be kept in mind that we should always draw the FBD of the body before solving any force questions. There are three mathematical quantities that will be of primary interest to us as we analyse the motion of objects in circles. These three quantities are speed, acceleration and force. In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. We should never forget to take the acceleration as the net acceleration before putting it to the equation.