A 4.3 g mass is released from rest at C which has a height of 0.8 m above the base of a loop-the-loop and a radius of 0.31 m. The acceleration of gravity is 9.8 m/s2 . Find the normal force pressing on the track at A, where A is at the same level as the center
Elena is correct.
PE=KE+PE₁
mgH = mv²/2 + mgR
v²= 2g(H-R)
at point A
N=mv²/R =2mg(H-R)/R=
=2•4.3 •10⁻³•9.8•(0.8-0.31)/0.31 =
= 0.133 N
To find the normal force pressing on the track at point A, we need to consider the different forces acting on the mass at that point.
1. Gravitational force: The weight of the mass exerts a downward force given by the formula F_g = m * g, where m is the mass and g is the acceleration due to gravity. In this case, m = 4.3 g (or 0.0043 kg) and g = 9.8 m/s^2.
F_g = (0.0043 kg) * (9.8 m/s^2) = 0.04214 N
2. Centripetal force: At point A, the mass is moving in a circular path with a radius of 0.31 m. To maintain this circular motion, there must be a centripetal force acting inward. The centripetal force is given by the formula F_c = m * a_c, where a_c is the centripetal acceleration.
The centripetal acceleration is related to the radius and tangential speed of the mass at point A. Since the mass is released from rest, the tangential speed at A can be found using the conservation of energy.
The potential energy at C is converted to kinetic energy at A. Therefore, we can write the equation:
m * g * h = (1/2) * m * v_A^2
Where h is the height of C above the base (0.8 m) and v_A is the tangential speed at A.
Solving for v_A:
v_A = √(2 * g * h)
v_A = √(2 * 9.8 m/s^2 * 0.8 m) = √(15.68 m^2/s^2) = 3.96 m/s
Now we can find the centripetal acceleration at A:
a_c = v_A^2 / r
Where r is the radius of the loop (0.31 m).
a_c = (3.96 m/s)^2 / 0.31 m = 50.674 m/s^2
Now we can calculate the centripetal force:
F_c = m * a_c
F_c = (0.0043 kg) * (50.674 m/s^2)
F_c = 0.21824 N
3. Normal force: To find the normal force pressing on the track at point A, we need to subtract the centripetal force from the gravitational force.
Normal force at A = F_g - F_c
Normal force at A = 0.04214 N - 0.21824 N
Normal force at A = -0.1761 N
Note that the negative sign indicates that the normal force is pointing downward. This negative value means that the track must exert an upward force on the mass to keep it moving in a circular path. Therefore, the magnitude of the normal force at A is 0.1761 N, and its direction is upward.
To find the normal force pressing on the track at point A, we need to consider the forces acting on the mass at that point. The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. In this case, the normal force will be directed towards the center of the loop-the-loop.
The mass is released from rest at point C, so initially, it only has gravitational potential energy. As it slides down from C to A, the potential energy is converted into kinetic energy. At point A, all the potential energy has been converted into kinetic energy, and the mass is moving horizontally. There is no vertical motion at point A, so the normal force will balance the weight of the mass.
To find the normal force, we can equate the weight of the mass to the centripetal force acting on it at point A. The centripetal force is the net force required to keep an object moving in a circular path.
First, let's find the speed of the mass at point A. We can use the conservation of energy principle:
Potential energy at point C = Kinetic energy at point A
mgh = (1/2)mv^2
where m is the mass, g is the acceleration due to gravity, h is the height, and v is the velocity.
Substituting the given values:
(4.3 g) * (9.8 m/s^2) * (0.8 m) = (1/2) * (4.3 g) * v^2
Simplifying the equation, we find:
v^2 = 2 * (9.8 m/s^2) * (0.8 m)
v^2 = 15.68 m^2/s^2
v = √(15.68 m^2/s^2)
v ≈ 3.96 m/s
Now, let's find the centripetal force at point A. The centripetal force is given by the equation:
Centripetal force = (mass) * (velocity)^2 / (radius)
F_c = (4.3 g) * (3.96 m/s)^2 / (0.31 m)
F_c = (0.043 kg) * (15.6816 m^2/s^2) / (0.31 m)
F_c = 2.184 N
Since the normal force at point A balances the weight of the mass vertically, it can be calculated as:
Normal force = Weight of the mass = (mass) * (gravity)
Normal force = (4.3 g) * (9.8 m/s^2)
Normal force = 42.14 N
Therefore, the normal force pressing on the track at point A is approximately 42.14 N.