A cyclist in an Olympic race is moving in a circular track of radius 80 m with speed 72 km/h. He has to lean from the vertical approximately through an angle of...
ANS: tan-1(1/2)
How is tan-1(1/2) derived in this answer?
To derive the angle at which the cyclist needs to lean from the vertical, we can use the concept of centripetal force in circular motion.
First, let's convert the speed of the cyclist from km/h to m/s:
72 km/h = (72 * 1000 m) / (3600 s) = 20 m/s
The centripetal force required to keep the cyclist moving in a circular path is provided by the friction between the bike tires and the track. This centripetal force is given by the formula:
F = mv^2/r
Where:
F = Centripetal force
m = Mass of the cyclist
v = Velocity of the cyclist
r = Radius of the circular track
We can rearrange this equation to solve for the angle at which the cyclist leans from the vertical (θ):
F = mg tan(θ)
Where:
g = Acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the expression for centripetal force (mv^2/r) into the equation:
mv^2/r = mg tan(θ)
Now, let's cancel out the mass (m) on both sides of the equation:
v^2/r = g tan(θ)
Substitute the given values:
(20 m/s)^2 / 80 m = 9.8 m/s^2 * tan(θ)
Simplify the equation:
tan(θ) = (20^2)/(80 * 9.8)
tan(θ) = 1/2
To find the angle θ, we take the inverse tangent (tan^-1) of both sides of the equation:
θ = tan^-1(1/2)
Therefore, the cyclist needs to lean from the vertical approximately through an angle of tan^-1(1/2).
To determine the angle that the cyclist needs to lean from the vertical in this scenario, we can use the concept of centripetal force. Centripetal force is the force that keeps an object moving in a circular path. In this case, the centripetal force is provided by the friction between the tires of the bicycle and the track.
We can start by calculating the centripetal acceleration (a) of the cyclist. The centripetal acceleration is given by the equation:
a = v^2 / r
where v is the linear velocity of the cyclist and r is the radius of the circular track.
Given that the speed of the cyclist is 72 km/h, we need to convert it to meters per second by dividing it by 3.6 (since 1 km/h = 1/3.6 m/s). Thus, the speed is:
v = 72 km/h ÷ 3.6 = 20 m/s
Substituting the values of v = 20 m/s and r = 80 m into the equation for centripetal acceleration, we get:
a = (20 m/s)^2 / 80 m = 5 m/s^2
Now, we need to find the angle at which the cyclist leans from the vertical. This angle can be calculated using the tangent function. The formula for the tangent of an angle (θ) is:
tan(θ) = opposite / adjacent
In this case, the opposite side is the centripetal acceleration (a), and the adjacent side is the gravitational acceleration (g).
Since the cyclist is moving in a circular path, the gravitational force acting on the cyclist can be considered as the weight of the cyclist (mg), where m is the mass of the cyclist and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Setting up our equation, we have:
tan(θ) = a / g
Substituting the values of a = 5 m/s^2 and g ≈ 9.8 m/s^2, we get:
tan(θ) = 5 / 9.8
To find θ, we use the inverse tangent function (tan^-1) on both sides of the equation:
θ = tan^-1(5 / 9.8)
Evaluating the expression on the right side, we get:
θ ≈ 0.515 radians
Hence, the cyclist has to lean from the vertical approximately through an angle of tan^-1(1/2), which is approximately 0.515 radians.
draw a sketch of the cyclist.
notice from her center of gravity, a horizontal force (centripetal force) of mv^2/r
the vertical force is mg
so tan Theta= mv^2/mgr=v^2/gr
now, the numbers v=72km/hr=20m/s (how convenient ) , and r=80
tanTheta=20^2/9.8*80=.51
theta=arctan(.51)