In a recent study of how mice negotiate turns, the mice ran around a circular 90 degree turn on a track with a radius of 0.15 m. The maximum speed measured for a mouse (mass = 18.5 g) running around this turn was 1.29 m/s. What is the minimum coefficient of friction between the track and the mouse’s feet that would allow a turn at this speed?
36.4
Well, if the mice are running so fast, they might as well join the circus! Now, onto the question. To find the minimum coefficient of friction, we need to consider the forces at play here.
In this case, the centripetal force is provided by the friction between the track and the mouse's feet. The formula for centripetal force is:
Fc = (m * v^2) / r
Where:
Fc is the centripetal force
m is the mass of the mouse
v is the velocity of the mouse
r is the radius of the turn
We can rearrange the formula to solve for the coefficient of friction:
μ = (m * v^2) / (r * g)
Where:
μ is the coefficient of friction
m is the mass of the mouse
v is the velocity of the mouse
r is the radius of the turn
g is the acceleration due to gravity
Now, let's plug in the values and calculate the coefficient of friction.
μ = (0.0185 kg * (1.29 m/s)^2) / (0.15 m * 9.8 m/s^2)
Solving this equation gives us the minimum coefficient of friction required for the mouse to make the turn at this speed. And remember, no clowning around with the math!
To determine the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed, we can use the following steps:
Step 1: Identify the relevant equations.
The centripetal force required to keep an object moving in a circular path is given by the equation:
Fc = (m * v^2) / r
Step 2: Convert the mass of the mouse to kilograms.
Given: mass of the mouse = 18.5 g
Convert grams to kilograms: 18.5 g * (1 kg / 1000 g) = 0.0185 kg
Step 3: Substitute the values into the equation.
Given: radius, r = 0.15 m
Given: maximum speed, v = 1.29 m/s
Given: mass, m = 0.0185 kg
Fc = (m * v^2) / r
= (0.0185 kg * (1.29 m/s)^2) / (0.15 m)
Step 4: Calculate the centripetal force required.
Fc = (0.0185 kg * 1.6641 m^2/s^2) / 0.15 m
= 0.20342 kg * m/s^2 / 0.15 m
= 1.3555 kg * m/s^2
Step 5: Assume that the maximum frictional force is equal to the centripetal force (since this is the minimum required for the turn).
Ff = Fc
Step 6: Determine the normal force acting on the mouse.
The normal force (Fn) is equal to the weight (mg) of the mouse when no vertical acceleration is present. Let's assume the acceleration of gravity (g) is approximately 9.8 m/s^2.
Fn = mg
= 0.0185 kg * 9.8 m/s^2
= 0.1803 kg * m/s^2
Step 7: Calculate the coefficient of friction.
The coefficient of friction (μ) is the ratio between the frictional force (Ff) and the normal force (Fn).
μ = Ff / Fn
= (1.3555 kg * m/s^2) / (0.1803 kg * m/s^2)
≈ 7.51
Therefore, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed is approximately 7.51.
To solve this problem, we need to consider the forces acting on the mouse as it runs around the turn. The two main forces are the gravitational force pulling the mouse downward and the frictional force between the mouse's feet and the track. The frictional force plays a crucial role in allowing the mouse to negotiate the turn without sliding.
The centripetal force is required to keep the mouse moving in a circular path. It is given by the equation:
F_centripetal = (mass × velocity^2) / radius
In this case, the mass of the mouse is 18.5 g, which is equivalent to 0.0185 kg. The velocity is given as 1.29 m/s, and the radius of the circular track is 0.15 m. Substituting these values into the equation, we get:
F_centripetal = (0.0185 kg × (1.29 m/s)^2) / 0.15 m
Now we need to find the minimum coefficient of friction (μ) that allows the mouse to navigate the turn at this speed. The frictional force (F_friction) is given by:
F_friction = μ × F_normal
The normal force (F_normal) is the force exerted by the track on the mouse's feet, which is equal to the gravitational force:
F_normal = mass × gravity
To find the minimum coefficient of friction, we can relate the frictional force to the centripetal force:
F_friction = F_centripetal
μ × F_normal = F_centripetal
μ × (mass × gravity) = (mass × velocity^2) / radius
Now we can substitute the known values into this equation:
μ × (0.0185 kg × 9.8 m/s^2) = (0.0185 kg × (1.29 m/s)^2) / 0.15 m
Simplifying this equation, we can solve for the coefficient of friction (μ):
μ = ((0.0185 kg × (1.29 m/s)^2) / 0.15 m) / (0.0185 kg × 9.8 m/s^2)
Evaluating this expression, we find:
μ = 0.99
Therefore, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed is approximately 0.99.