A 9.97kg object slides along the frictionless loop the loop with a diameter of 3 m . what minimum speed must it have at the loop in order to remain in contact with the loop ?
To determine the minimum speed required for the object to remain in contact with the loop, we can use the principle of conservation of mechanical energy.
The object will experience two main forces as it goes around the loop: the force of gravity (mg) and the normal force (N) exerted by the loop on the object. At the minimum speed required, the normal force will be zero at the top of the loop.
At the top of the loop, the object will momentarily lose contact with the loop when the normal force becomes zero. In this case, the gravitational force will be solely responsible for keeping the object moving in a circular path.
The gravitational potential energy at the bottom of the loop will be converted to kinetic energy at the top of the loop, since there is no friction or other dissipative forces involved.
Using the conservation of mechanical energy equation:
Initial Energy (at the bottom) = Final Energy (at the top)
mgh (initial) = 0.5 * m * v^2 (final)
where m is the mass of the object, g is the acceleration due to gravity, h is the height of the loop, and v is the minimum speed required.
In this case, the height of the loop is the radius of the loop, which is half the diameter:
h = r = 3m / 2 = 1.5m
Substituting the values into the equation:
9.97kg * 9.8m/s^2 * 1.5m = 0.5 * 9.97kg * v^2
145.089J = 4.985 * v^2
Dividing both sides by 4.985:
29.1 = v^2
Taking the square root of both sides:
v ≈ 5.4 m/s
Therefore, the minimum speed required for the object to remain in contact with the loop is approximately 5.4 m/s.
To find the minimum speed the object must have at the loop in order to remain in contact, we can use the concept of centripetal force.
First, let's calculate the gravitational force acting on the object while it's at the top of the loop. The object's weight can be determined using the formula:
F = m * g
Where:
F is the gravitational force (weight),
m is the mass of the object,
g is the acceleration due to gravity (approximately 9.8 m/s²).
F = 9.97 kg * 9.8 m/s²
F = 97.7066 N
Next, we need to find the minimum centripetal force required for the object to remain in contact with the loop. At the top of the loop, the centripetal force is provided by the normal force acting on the object.
At the top of the loop, the object experiences two forces: weight (gravity), acting downward, and the normal force, acting upward. The difference between these two forces provides the net centripetal force required.
The normal force can be determined using the equation:
F_normal = F_gravity + F_net
Where:
F_normal is the normal force,
F_gravity is the gravitational force (weight),
F_net is the net centripetal force.
Since there is no friction, the net force is provided solely by the normal force.
So, we can rewrite the equation as:
F_normal = F_gravity + F_centripetal
Since the object is at the top of the loop, the normal force (F_normal) is equal to the centripetal force (F_centripetal).
F_centripetal = F_gravity
Now we can substitute the gravitational force into the equation to solve for the minimum centripetal force:
F_centripetal = 97.7066 N
The centripetal force is given by the formula:
F_centripetal = m * v² / r
Where:
F_centripetal is the centripetal force,
m is the mass of the object,
v is the velocity,
r is the radius of the loop (diameter divided by 2).
We know the mass of the object (m = 9.97 kg) and the radius of the loop (r = 3 m). We need to solve for v.
Rearranging the formula:
v² = F_centripetal * r / m
v² = 97.7066 N * 3 m / 9.97 kg
v² = 29.1022 N.m / 9.97 kg
v² ≈ 2.9190 m²/s²
Finally, taking the square root of both sides to find v:
v ≈ √2.9190 m²/s²
v ≈ 1.71 m/s
Therefore, the minimum speed the object must have at the loop in order to remain in contact with the loop is approximately 1.71 m/s.