The flat circular disc rotates about a vertical axis through O with a constant angular velocity of 240rpm. Prior to rotation, each of the 0.5kg sliding blocks has the position x=25mm with no force in its attached spring. Each spring has a stiffness of 400N/m, Neglect any friction between the blocks and the slots, and neglect the mass of the springs.
(a) Determine the value of x for each spring
(b) Calculate the normal force N exerted by the side of the slot on the block
(a) Well, let's go on a springy journey, shall we? Since the angular velocity is constant, the centripetal acceleration of each block will be constant as well. This means that the net force on each block will also be constant. So, let's find that net force!
The displacement of each block from its equilibrium position is given by x = 25mm. The force exerted by each spring can be found using Hooke's Law: F = kx, where F is the force, k is the spring constant (stiffness), and x is the displacement.
So, for each block, the force exerted by the spring is F = 400N/m * 0.025m = 10N. And since the net force is constant, it must be equal to this force. Therefore, the value of x for each spring is 0.025m, or 25mm.
(b) Now, let's tackle the normal force exerted by the side of the slot on the block. Since there's no friction between the blocks and the slots, the only force acting on the block in the vertical direction is the normal force, denoted by N.
To find N, we need to consider the forces acting on the block. In the vertical direction, we have the weight of the block acting downwards, given by mg, where m is the mass of the block. Since the block is sliding, its weight does not contribute to the normal force.
So, the normal force N is equal to the net force acting in the vertical direction, which in this case is simply the force exerted by the spring, 10N.
Therefore, the normal force N exerted by the side of the slot on the block is 10N. And don't worry, it won't make the block feel "abnormal"!
To determine the value of x for each spring, we need to consider the equilibrium condition for the sliding blocks at the given angular velocity.
(a) The equilibrium condition for each sliding block can be expressed as the balance between the gravitational force and the spring force. Since there is no friction, the normal force and the centripetal force can be disregarded.
First, let's convert the angular velocity from rpm to rad/s:
ω = (240 rpm * 2π rad/1 min) / (60 s) = 8π rad/s
We can assume that the gravitational force acting on each block is equal to its weight, which can be calculated as:
F_gravity = m * g = 0.5 kg * 9.8 m/s² = 4.9 N
Next, we need to calculate the spring force. The spring force is given by Hooke's Law: F_spring = -k * x, where k is the spring stiffness and x is the displacement from the equilibrium position.
Since the block is in equilibrium, the spring force and the gravitational force are equal in magnitude:
|F_spring| = |F_gravity|
Substituting the values, we get:
k * x = 4.9 N
Now, we can solve for x:
x = 4.9 N / k
Substituting the given spring stiffness (k = 400 N/m), we calculate:
x = 4.9 N / 400 N/m
x = 0.01225 m = 12.25 mm
Therefore, the value of x for each spring is 12.25 mm.
(b) To calculate the normal force N exerted by the side of the slot on the block, we can use the centripetal force acting on the block.
The centripetal force can be calculated as the product of mass and centripetal acceleration:
F_centripetal = m * a_c
The centripetal acceleration can be obtained using the formula:
a_c = r * ω^2
Given that the block is rotating in a circular path of radius r = 25 mm = 0.025 m and angular velocity ω = 8π rad/s, we can substitute these values into the centripetal acceleration equation:
a_c = 0.025 m * (8π rad/s)^2
Calculating the centripetal acceleration, we find:
a_c = 16π^2 m/s²
Therefore, the centripetal force is:
F_centripetal = 0.5 kg * 16π^2 m/s²
Since there is no other force acting on the block in the vertical direction, the normal force N is equal in magnitude but opposite in direction to the centripetal force:
|N| = |F_centripetal| = 0.5 kg * 16π^2 m/s²
Calculating the normal force, we have:
N = 0.5 kg * 16π^2 m/s²
Hence, the normal force exerted by the side of the slot on the block is equal to 8π^2 N.