A solid ball is released from rest and slides down a hillside that slopes downward at an angle 52.0 from the horizontal.What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur?
To find the minimum value of the coefficient of static friction between the hill and ball surfaces required for no slipping to occur, we need to consider the forces acting on the ball.
First, let's draw a free-body diagram of the ball. The weight of the ball (mg) acts vertically downwards, and there are two forces acting parallel to the inclined surface of the hill: the normal force (N) exerted by the hill on the ball, and the friction force (f) opposing the motion of the ball. We also have the angle between the hill and the horizontal, which is given as 52.0°.
Now, let's resolve the weight force into its components. The component of the weight force parallel to the hill is mg*sin(52.0°), and the component perpendicular to the hill is mg*cos(52.0°).
Since we want to find the minimum value of the coefficient of static friction for no slipping, the maximum force of static friction (fs) should be equal to the component of the weight force parallel to the hill.
fs = mg*sin(52.0°)
According to the laws of static friction, the maximum force of static friction is given by:
fs = μs * N
where μs is the coefficient of static friction, and N is the normal force. The normal force is equal to the component of the weight force perpendicular to the hill:
N = mg*cos(52.0°)
Substituting this value of N in the equation for fs, we have:
mg*sin(52.0°) = μs * mg*cos(52.0°)
Simplifying the equation, we can cancel out the mass (m) on both sides:
sin(52.0°) = μs * cos(52.0°)
Dividing both sides by cos(52.0°), we get:
tan(52.0°) = μs
Hence, the minimum value of the coefficient of static friction required for no slipping to occur is given by the tangent of the angle between the hill and the horizontal, which is 52.0°.
To determine the minimum value of the coefficient of static friction required for no slipping to occur, we can analyze the forces acting on the ball as it slides down the hillside.
Let's consider the forces in the horizontal direction (along the slope) and the vertical direction (perpendicular to the slope).
In the horizontal direction:
1. The gravitational force component parallel to the slope is mg * sin(52.0°), where m is the mass of the ball and g is the acceleration due to gravity.
2. The static friction force opposes the gravitational force component along the slope. So, the static friction force is given by fs ≤ μ * N, where μ is the coefficient of static friction and N is the normal force exerted by the hill on the ball.
In the vertical direction:
1. The gravitational force component perpendicular to the slope is mg * cos(52.0°).
2. The normal force N is equal to this gravitational force component perpendicular to the slope. So, N = mg * cos(52.0°).
To prevent slipping, the static friction force must be equal to or greater than the gravitational force component along the slope. Therefore, we have:
fs ≥ mg * sin(52.0°)
Substituting the expression for fs from above, we get:
μ * N ≥ mg * sin(52.0°)
Now, substituting the value of N, we have:
μ * mg * cos(52.0°) ≥ mg * sin(52.0°)
The mass cancels out, giving us:
μ * cos(52.0°) ≥ sin(52.0°)
Finally, dividing both sides by cos(52.0°), we find:
μ ≥ tan(52.0°)
Therefore, the minimum value for the coefficient of static friction between the hill and ball surfaces to prevent slipping is tan(52.0°).
Well, let me calculate that for you. Ok, carry the bacon, divide by the square root of pi... and drumroll please... the minimum coefficient of static friction needed for no slipping to occur is the same as the minimum amount of coffee needed for me to function in the morning. In other words, it's quite important!
Let u be the static friction foefficient.
Linear Acceleration = a
= M[g*sin52/M -M*g cos52*u]/M
= g*[sin52 - u cos52]
Angular acceleration = alpha = a/R
= g [sin52 -ucos52)/R
(if there is no slipping)
= Ff*R/I < u*M*g*sin52*R/[(2/5)M R^2]
= (5/2)*u*g*sin52/R
Ff is the static friction force, which has a maximum value when slipping occurs.
To avoid slipping for a solid sphere, at that 52 degree slope angle,
(5/2)*u*g*sin52/R > g [sin52 -ucos52)/R
g's and R's cancel.
u*[(5/2)sin52 + cos52] > sin 52
u > 1/[(5/2) + cot 52] = 0.305