Centripetal Force/Changing direction on ice - Turn length

[testme]

Homework Statement

I am running to school at 6m/s when I hit a patch of icy sidealk. The coefficient of friction between the ice and my shoes is 0.16. I wish to turn my 55kg frame to the left. What is the smallest turn I can make?

v = 6 m/s
μ = 0.16
m = 55 kg

Homework Equations

Fnet = mv^2/r
Ff = μFn
Fnet = ma

The Attempt at a Solution

Assume that up and left are positive.

Fnet = ma
Fg + Fn = ma
-539 + Fn = 0
Fn = 539

Ff = μFn
Ff = 0.16 (539)
Ff = 86.24

Fnet = mv^2/r
Ff = mv^2/r
86.24 = 55(6)^2/r
86.24 = 1980/r
r = 23 m

I'm not sure if that's the answer or what to do next, or even if what I did is the right method. The question is kind of confusing me because I don't fully understand how the person is moving.

 

[ehild]

r = 23 m

I'm not sure if that's the answer or what to do next, or even if what I did is the right method. The question is kind of confusing me because I don't fully understand how the person is moving.

You got the radius of the circle (correct) the boy can turn without slipping outwards. He wants to turn to the left, change direction by 90 degrees. Instead of just turning round his body, he makes the turn along an arc. I think the problem asks the length of the arc.

ehild

 

[testme]

So you mean like a quarter of the circumference?

C = 2∏r
C = 46∏
C = 145
C/4 = 36.25m

Therefore the smallest turn the boy can make is 36.25m?

 

[ehild]

So you mean like a quarter of the circumference?

C = 2∏r
C = 46∏
C = 145
C/4 = 36.25m

Therefore the smallest turn the boy can make is 36.25m?

That is, (pi/2)*R. . Yes, I think, that was the question.

ehild

 

[asteriatic]

I would like to clarify a few things in order to provide a more accurate response. First, it is important to specify the direction in which the person is initially running. Is it in a straight line or at an angle? This will affect the direction in which they need to turn. Additionally, the statement mentions a coefficient of friction between the ice and the shoes, but it does not specify the type of shoes being worn. Different types of shoes may have different coefficients of friction on ice.

Assuming that the person is initially running in a straight line and is able to turn their body without any external forces acting on them, we can calculate the minimum turn length using the centripetal force equation:

Fnet = mv^2/r

Where Fnet is the net force acting on the person, m is their mass, v is their initial velocity, and r is the radius of the turn. In this scenario, the net force is equal to the force of friction, Ff, which is given by Ff = μFn, where μ is the coefficient of friction and Fn is the normal force. The normal force can be calculated by balancing the forces in the vertical direction, which would give us Fn = mg, where g is the acceleration due to gravity.

Substituting these values into the centripetal force equation, we get:

Ff = mv^2/r
μFn = mv^2/r
μmg = mv^2/r
r = μmg/v^2

Plugging in the given values, we get:

r = (0.16)(55 kg)(9.8 m/s^2)/(6 m/s)^2
r = 10.78 m

Therefore, the minimum turn length on the icy sidewalk is approximately 10.78 meters. This assumes that the person is able to turn their body without any external forces acting on them, which may not be the case in a real-world scenario. Other factors such as the person's balance and coordination, the type of shoes being worn, and the surface of the ice may also affect the minimum turn length.