A uniform rod of mass M = 0.245kg and length L = 0.49m stands vertically on a horizontal table. It is released from rest to fall.
(a) Calculate the angular speed of the rod, the vertical acceleration of the moving end of the rod, and the normal force exerted by the table on the rod as it makes an angle θ = 48.8° with respect to the vertical.
(b) If the rod falls onto the table without slipping, find the linear acceleration of the end point of the rod when it hits the table.
Please help, Thanks!
√3q/l
To solve this problem, we can use principles of rotational motion and linear kinematics. Let's break it down step by step:
(a) Calculating the angular speed of the rod:
To find the angular speed (ω) of the rod, we need to apply principles of rotational motion. Since the rod is released from rest, we can use the principle of conservation of energy.
The potential energy at the initial position is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the initial height.
The initial height of the rod can be calculated using the formula h = L(1 - cosθ), where L is the length of the rod and θ is the angle made with the vertical.
Therefore, the initial potential energy is PE = mgL(1 - cosθ).
At the final position, when the rod makes an angle θ with respect to the vertical, all the potential energy is converted to rotational kinetic energy. The rotational kinetic energy (KE) is given by the formula KE = (1/2)Iω², where I is the moment of inertia and ω is the angular speed.
The moment of inertia of a uniform rod about its axis of rotation is given by the formula I = (1/3)mL².
Equating the potential energy and rotational kinetic energy, we have:
mgL(1 - cosθ) = (1/2)(1/3)mL²ω².
Simplifying the equation, we get:
ω² = (6g(1 - cosθ))/L.
Substituting the given values:
ω² = (6 * 9.8(1 - cos48.8°))/0.49.
Solving this equation, we find:
ω ≈ 13.33 rad/s.
(b) Calculating the linear acceleration of the end point of the rod:
Since the rod is falling without slipping, the linear acceleration of the end point is related to the angular acceleration (α) by the formula α = a/R, where a is the linear acceleration and R is the radius of the rod.
The radius of the rod is given by R = L/2.
We can also relate the angular acceleration to the angular speed using the formula α = ω²/R.
Combining these equations, we have:
a/R = ω²/R.
Simplifying, we get:
a = ω²R = ω²(L/2).
Substituting the given value of ω, we have:
a = (13.33 rad/s)²(0.49m/2).
Solving this equation, we find:
a ≈ 17.45 m/s².
(c) Calculating the normal force exerted by the table on the rod:
To find the normal force (N) exerted by the table on the rod, we need to consider the forces acting on the rod. There are two forces: the weight of the rod acting vertically downward (mg), and the normal force exerted by the table acting vertically upward (N).
In the vertical direction, the net force is given by:
ma = mg - N.
Since the rod is accelerating downward, the net force is in the downward direction. Therefore, the equation becomes:
ma = mg - N.
Substituting the given values:
0.245a = 0.245 * 9.8 - N.
Simplifying, we get:
N = 0.245 * 9.8 - 0.245a.
Substituting the value of a, we find:
N ≈ 4.78 N.
So, the normal force exerted by the table on the rod is approximately 4.78 Newtons.
Therefore, the answers to the given questions are as follows:
(a) The angular speed of the rod is approximately 13.33 rad/s.
(b) The linear acceleration of the end point of the rod is approximately 17.45 m/s².
(c) The normal force exerted by the table on the rod is approximately 4.78 N.