a single bead can slide with negligible friction on a wire that is bent into circular loop of radius 15 cm an in the figure. the circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.450 s. the position of the bead is described by the angle t that the radial line,from the center of the loop to thebead,makes with the vertical.at what angle up from the bottom of the circlecan the bead stay motionless relative to the turning circle?
Well, let me put on my clown nose and see if I can give you a laugh while helping out with your question!
Now, imagine the bead at the bottom of the circle. It's like a daredevil balancing act on a unicycle! The bead wants to stay motionless relative to the turning circle, so it needs to find the perfect angle to keep its balance.
Since we are dealing with a circular motion, we know that there must be a centripetal force acting towards the center of the circle. In this case, the tension in the wire is providing that force.
At the bottom of the circle, the tension has to counterbalance the weight of the bead. The tension force is equal to the weight of the bead to keep it motionless. So, we can set up an equation:
Tension = M * g,
where M is the mass of the bead and g is the acceleration due to gravity.
But we also know that the tension force can be expressed as:
Tension = M * a_c,
where a_c is the centripetal acceleration.
Now, the centripetal acceleration is given by:
a_c = ω^2 * r,
where ω is the angular velocity and r is the radius of the circle.
Since we are looking for the angle relative to the bottom of the circle, we need to find the smallest angle where the centripetal acceleration is canceled out by the weight of the bead. That angle occurs when the bead is at the highest point of the circle.
So, plug in the values into the equations, solve for the angle, and you'll find the answer!
To find the angle at which the bead can stay motionless relative to the turning circle, we need to consider the forces acting on the bead.
The two main forces acting on the bead are gravity and the normal force. The normal force acts perpendicular to the wire, while gravity acts vertically downwards.
At the bottom of the circle, the normal force is at its maximum value, which is equal to the sum of the weight of the bead and the centripetal force required to keep the bead moving in a circle.
At the top of the circle, the normal force is at its minimum value, which is equal to the difference between the weight of the bead and the centripetal force required to keep the bead moving in a circle.
At the point where the bead can stay motionless relative to the turning circle, the normal force is equal to zero.
Let's consider the forces acting on the bead at this point:
1. Weight (mg) acting vertically downwards
2. Centripetal force (mv^2/r) acting radially towards the center of the circle
At the point where the bead can stay motionless relative to the turning circle, the weight and centripetal force combine to create a net force of zero in the radial direction. This means that the weight is balanced by the centripetal force.
Mathematically, we can write:
mg = mv^2/r
Cancelling the mass (m) from both sides of the equation gives:
g = v^2/r
We can then find the velocity (v) of the bead using the formula:
v = 2πr/T
Where:
- v is the velocity of the bead
- r is the radius of the circle
- T is the period of rotation
Substituting this into the equation above gives:
g = (2πr/T)^2 / r
Simplifying further:
g = (4π^2r) / T^2
Now, we can solve for the angle (θ) at which the bead can stay motionless relative to the turning circle:
tan(θ) = r / (2πr) * g / T^2
tan(θ) = 1 / (2π) * g / T^2
θ = arctan(1 / (2π) * g / T^2)
Plugging in the values for g, r, and T from the given information:
g = 9.8 m/s^2
r = 15 cm = 0.15 m
T = 0.450 s
θ = arctan(1 / (2π) * 9.8 m/s^2 / (0.450 s)^2)
Evaluating this expression gives the angle (θ) at which the bead can stay motionless relative to the turning circle.
To find the angle at which the bead can stay motionless relative to the turning circle, we need to consider the forces acting on the bead.
The bead experiences two forces: gravity, which pulls it downwards, and the frictional force from the circular motion of the loop. The frictional force needs to balance the gravitational force for the bead to stay motionless.
The gravitational force can be calculated using the formula: F_gravity = m * g, where m is the mass of the bead and g is the acceleration due to gravity.
The frictional force depends on the centripetal force required for the circular motion. Centripetal force can be calculated using the formula: F_centripetal = m * ω^2 * r, where m is the mass of the bead, ω is the angular velocity (ω = 2π / T, where T is the period of rotation), and r is the radius of the circular loop.
Since the bead is motionless, we can equate the gravitational force to the centripetal force:
m * g = m * ω^2 * r
The mass (m) cancels out from both sides of the equation:
g = ω^2 * r
Now, let's solve for the angle at which the bead can stay motionless. The angle can be found using the formula:
sin(t) = r / R
Where R is the radius of the circle. Given that R = 15 cm, we can substitute the value of r in terms of R:
sin(t) = (15 cm) / R
To find the value of t, we can take the inverse sine (or arcsin) of both sides:
t = sin^(-1)((15 cm) / R)
Now, we can substitute the given value of R (15 cm) to find the angle t.