A coin with mass 3.1 grams is at rest atop a wooden block of mass 20 grams. The block is on a spinning platform with a 12 centimeter radius.
The coefficients of friction between the platform and the wooden block are μs = 0.75 and μk = 0.64
The coefficient of friction between the wooden block and the coin are μs = 0.52 and μk = 0.45
What is the maximum speed the penny and block can have so that neither of them slide off of the spinning platform? Convert that answer into revolutions per minute to get a rate of rotation.
To find the maximum speed at which the penny and block can have without sliding off the platform, we need to consider the forces acting on them and the frictional forces.
1. Find the maximum frictional force between the wooden block and the spinning platform:
The maximum static frictional force is given by the equation:
F_s = μs * N
where F_s is the maximum static frictional force, μs is the coefficient of static friction, and N is the normal force.
The normal force N can be calculated as:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity.
Substituting the given values:
N = (20 grams) * (9.8 m/s^2) = 0.196 N
F_s = (0.75) * (0.196 N) = 0.147 N
2. Find the maximum frictional force between the coin and the wooden block:
Using the same process as above, but with the mass of the coin (3.1 grams) and the coefficient of static friction (0.52):
N = (3.1 grams) * (9.8 m/s^2) = 0.030 N
F_s = (0.52) * (0.030 N) = 0.016 N
3. Calculate the net force acting on the coin-block system:
The net force is the difference between the maximum static frictional force between the block and the platform and the maximum static frictional force between the coin and the block:
Net force = F_s(block-platform) - F_s(coin-block)
Net force = 0.147 N - 0.016 N = 0.131 N
4. Find the maximum centripetal force that the spinning platform can exert to prevent the coin-block system from sliding off:
The centripetal force is given by the equation:
F_c = m * (v^2 / r)
where F_c is the centripetal force, m is the total mass of the coin-block system, v is the velocity, and r is the radius of the spinning platform.
The total mass of the coin-block system is 20 grams + 3.1 grams = 23.1 grams = 0.0231 kg.
Rearranging the equation, we get:
v = sqrt(F_c * r / m)
5. Substitute the values and solve for the maximum velocity:
v = sqrt((0.131 N) * (0.12 m) / (0.0231 kg))
v = sqrt(0.004524 Nm/kg)
v = 0.0674 m/s
6. Finally, convert the maximum velocity to revolutions per minute:
To convert the velocity from meters per second to revolutions per minute, we need to multiply by a conversion factor.
1 m/s = 60 * (1 / (2π * r) RPM
v_rpm = 0.0674 m/s * 60 * (1 / (2π * 0.12 m) = 16.58 RPM
Therefore, the maximum speed the penny and block can have without sliding off of the spinning platform is approximately 0.0674 m/s or 16.58 RPM.
To find the maximum speed the penny and block can have without sliding off the spinning platform, we need to consider the forces acting on them.
First, let's calculate the force between the block and the platform due to the mass of the block. The force is given by:
F_block = mass_block * gravity
where mass_block is the mass of the wooden block and gravity is the acceleration due to gravity (approximately 9.8 m/s^2).
F_block = 20 g * 9.8 m/s^2
F_block = 196 g m/s^2
Now, let's calculate the maximum frictional force between the block and the platform. The maximum frictional force is given by:
F_max = μs * F_block
where μs is the coefficient of static friction between the block and the platform.
F_max = 0.75 * 196 g m/s^2
F_max = 147 g m/s^2
Next, let's calculate the force between the coin and the block due to the mass of the coin. The force is given by:
F_coin = mass_coin * gravity
where mass_coin is the mass of the coin.
F_coin = 3.1 g * 9.8 m/s^2
F_coin = 30.38 g m/s^2
Now, let's calculate the maximum frictional force between the coin and the block. The maximum frictional force is given by:
F_max_coin = μs * F_coin
where μs is the coefficient of static friction between the coin and the block.
F_max_coin = 0.52 * 30.38 g m/s^2
F_max_coin = 15.81 g m/s^2
Since the maximum frictional force between the block and the platform is greater than the maximum frictional force between the coin and the block (147 g m/s^2 > 15.81 g m/s^2), we need to consider the block sliding off the platform rather than the coin sliding off the block.
To prevent the block from sliding off, the centripetal force acting on it must be less than or equal to the maximum frictional force between the block and the platform. The centripetal force is given by:
F_centripetal = mass_block * acceleration_circular
where acceleration_circular is the centripetal acceleration of the block.
The centripetal acceleration is given by:
acceleration_circular = (velocity^2) / radius
where velocity is the speed of the block and coin, and radius is the radius of the platform.
To convert the speed to revolutions per minute (rpm), we need to divide the speed by the circumference of the circular path, which is given by:
circumference = 2 * π * radius
Now let's put it all together to find the maximum speed:
F_centripetal = mass_block * ((velocity^2) / radius)
F_centripetal = mass_block * (velocity^2) / 0.12
For the block not to slide off, the centripetal force must be less than or equal to the maximum frictional force:
mass_block * (velocity^2) / 0.12 ≤ 147 g m/s^2
Let's solve this equation for velocity:
velocity ≤ sqrt((147 g m/s^2 * 0.12) / mass_block)
Now, substitute the mass of the block into the equation:
velocity ≤ sqrt((147 g m/s^2 * 0.12) / 20 g)
velocity ≤ sqrt(0.0882 m^2/s^2)
velocity ≤ 0.297 m/s
To convert this velocity to revolutions per minute (rpm), we need to divide it by the circumference of the circular path:
circumference = 2 * π * 0.12 m
circumference ≈ 0.754 m
Now, let's convert the velocity to rpm:
velocity_rpm = velocity / circumference
velocity_rpm = 0.297 m/s / 0.754 m
velocity_rpm ≈ 0.393 rpm
Therefore, the maximum speed the penny and block can have without sliding off the spinning platform is approximately 0.393 rpm.