A 64.8 g drink is left on a turntable causing it to turn with an angular speed of 8.50 rad/s. If the drink has coefficients of friction of μs = 0.890 and μk = 0.500 with the turntable, what is the furthest distance that the drink could have been placed from the centre of the turntable for the drink not to slide?
To solve this problem, we need to consider the equilibrium condition for rotational motion. The maximum distance the drink can be placed from the center without sliding can be determined by balancing the gravitational force and the frictional force acting on the drink.
Let's break down the solution step by step:
Step 1: Determine the gravitational force acting on the drink.
The gravitational force (Fg) can be calculated using the formula:
Fg = m * g
where m is the mass of the drink and g is the acceleration due to gravity.
Given that the mass of the drink is 64.8 g (which is equivalent to 0.0648 kg) and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the gravitational force as follows:
Fg = (0.0648 kg) * (9.8 m/s²)
Fg ≈ 0.63504 N
Step 2: Determine the maximum static friction force.
The maximum static friction force (Fs) is equal to the coefficient of static friction (μs) multiplied by the normal force (N), which is the force exerted by the turntable on the drink.
The normal force can be calculated using the formula:
N = m * g
where m is the mass of the drink and g is the acceleration due to gravity.
So, the normal force is:
N = (0.0648 kg) * (9.8 m/s²)
N ≈ 0.63504 N
Now we can calculate the maximum static friction force:
Fs = μs * N
Fs = 0.890 * 0.63504 N
Fs ≈ 0.56523 N
Step 3: Determine the maximum distance (r) that the drink can be placed from the center without sliding.
The maximum distance (r) can be determined using the equation:
Fs = r * m * ω²
where Fs is the maximum static friction force, r is the distance from the center of the turntable to the drink, m is the mass of the drink, and ω is the angular speed of the turntable.
Now we can rearrange the equation to solve for r:
r = Fs / (m * ω²)
r = 0.56523 N / ((0.0648 kg) * (8.50 rad/s)²)
r ≈ 0.996 m
Therefore, the drink can be placed at a maximum distance of approximately 0.996 meters from the center of the turntable without sliding.
To find the furthest distance that the drink could have been placed from the center of the turntable for it not to slide, we need to consider the forces acting on the drink and determine the maximum force of static friction that can prevent the drink from sliding.
The force of static friction (F(static)) is given by the equation:
F(static) = μs * N
where μs is the coefficient of static friction and N is the normal force.
The normal force (N) is equal to the weight of the drink, which is given by:
N = m * g
where m is the mass of the drink and g is the acceleration due to gravity.
In this case, the mass of the drink is 64.8 g, which is equivalent to 0.0648 kg.
The acceleration due to gravity is approximately 9.8 m/s².
So, the normal force is:
N = 0.0648 kg * 9.8 m/s²
Next, we need to determine the maximum force of static friction (F(static)) that can be exerted to prevent the drink from sliding. This maximum force of static friction is given by:
F(max) = μs * N
where μs is the coefficient of static friction.
In this case, the coefficient of static friction is given as μs = 0.890.
So, the maximum force of static friction is:
F(max) = 0.890 * N
Finally, the furthest distance that the drink could have been placed from the center of the turntable without sliding can be calculated using the centripetal force equation:
F(c) = m * ω² * r
where F(c) is the centripetal force, m is the mass of the drink, ω is the angular speed of the turntable, and r is the distance from the center of the turntable.
In this case, we need to solve for r.
Rearranging the equation, we have:
r = F(c) / (m * ω²)
First, we need to determine the centripetal force (F(c)), which is equal to the force of static friction (F(static)):
F(c) = F(static) = F(max)
Now we can calculate the furthest distance:
r = F(max) / (m * ω²)
Substituting the values we have:
r = (0.890 * N) / (0.0648 kg * (8.50 rad/s)²)
After evaluating this equation, we can find the furthest distance that the drink could have been placed from the center of the turntable without sliding.
the mass is not a factor
(8.5 * r)^2 / r = .89 g
r = .89 g / (8.5^2)