
A cyclist is riding with a speed of $27km/hr$ As he approaches a circular turn on the road of radius 80m, he applies brakes and reduces his speed at the constant rate of $0.5m/s$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer
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Hint: This question is from the concept of circular motion which is part or mechanics where the acceleration of any object or vehicle performing circular motion is divided into two perpendicular components called tangential acceleration and normal acceleration (centripetal acceleration).
Complete step-by-step answer:
Given
The speed of the cyclist is $27km/hr = 7.5m/s$
Radius of the road is 80m
Retardation due to breaking is $0.5m/{s^2}$
As we know that the centripetal acceleration is given as ${a_c} = \dfrac{{{V^2}}}{r}$
Where a is acceleration, V is the velocity and r is the radius of the circular path
Now, using the above formula to find the centripetal acceleration
${a_c} = \dfrac{{{V^2}}}{r}$
Substituting the value of r and V in the above equation, we get
$
{a_c} = \dfrac{{{{\left( {7.5} \right)}^2}}}{{80}} \\
{a_c} = 0.70m/{s^2} \\
$
As we know that the centripetal acceleration and the tangential acceleration are at 90 degree to each other, therefore the resultant acceleration is given by
$a = {\left( {a_c^2 + a_T^2} \right)^{\dfrac{1}{2}}}$
$
a = \sqrt {{{0.7}^2} + {{0.5}^2}} \\
a = 0.86m/{s^2} \\
$
Hence, the resultant acceleration of the cyclist is $0.86m/{s^2}$
The direction of the net acceleration can be found by finding the angle between resultant and the tangential acceleration which is given by
$
\tan \theta = \dfrac{{{a_c}}}{{{a_T}}} \\
\tan \theta = \dfrac{{0.7}}{{0.5}} = 1.4 \\
\theta = {54.56^0} \\
$
The direction of the net acceleration of the cyclist on the circular turn is ${54.56^0}$
Note- As any vehicle like cycle makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. The acceleration is directed radially toward the center of the circle and has a magnitude equal to the square of the body's speed along the curve divided by the distance from the center of the circle to the moving body.
Complete step-by-step answer:
Given
The speed of the cyclist is $27km/hr = 7.5m/s$
Radius of the road is 80m
Retardation due to breaking is $0.5m/{s^2}$
As we know that the centripetal acceleration is given as ${a_c} = \dfrac{{{V^2}}}{r}$
Where a is acceleration, V is the velocity and r is the radius of the circular path
Now, using the above formula to find the centripetal acceleration
${a_c} = \dfrac{{{V^2}}}{r}$
Substituting the value of r and V in the above equation, we get
$
{a_c} = \dfrac{{{{\left( {7.5} \right)}^2}}}{{80}} \\
{a_c} = 0.70m/{s^2} \\
$
As we know that the centripetal acceleration and the tangential acceleration are at 90 degree to each other, therefore the resultant acceleration is given by
$a = {\left( {a_c^2 + a_T^2} \right)^{\dfrac{1}{2}}}$
$
a = \sqrt {{{0.7}^2} + {{0.5}^2}} \\
a = 0.86m/{s^2} \\
$
Hence, the resultant acceleration of the cyclist is $0.86m/{s^2}$
The direction of the net acceleration can be found by finding the angle between resultant and the tangential acceleration which is given by
$
\tan \theta = \dfrac{{{a_c}}}{{{a_T}}} \\
\tan \theta = \dfrac{{0.7}}{{0.5}} = 1.4 \\
\theta = {54.56^0} \\
$
The direction of the net acceleration of the cyclist on the circular turn is ${54.56^0}$
Note- As any vehicle like cycle makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. The acceleration is directed radially toward the center of the circle and has a magnitude equal to the square of the body's speed along the curve divided by the distance from the center of the circle to the moving body.
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