A small object of mass m= 90 kg slides down a spherical dome of radius R=12 m without any friction. It starts off at the top (polar angle θ=0) at zero speed. Use g=10 m/s2. (See figure)
(a) What is the magnitude of the force (in Newtons) exerted by the dome on the mass when it is at the top, at θ=0∘?
N(θ=0∘)=
(b) What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘?
N(θ=30∘)=
(c) At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘; )
θ0=
Midterm of 801.x?
Take down.
To solve this problem, we need to consider the forces acting on the small object as it slides down the spherical dome.
Let's first determine the force at the top of the dome (θ = 0°), which is part (a) of the question.
(a) When the object is at the top of the dome (θ = 0°), it experiences a gravitational force pulling it downward. The magnitude of this force can be calculated using the formula: F = mg, where m is the mass of the object and g is the acceleration due to gravity.
Given that the mass of the object is m = 90 kg and g = 10 m/s^2, we can find the force:
F(θ = 0°) = mg = 90 kg * 10 m/s^2 = 900 N
Therefore, the magnitude of the force exerted by the dome on the object at θ = 0° is N(θ = 0°) = 900 N.
Moving on to part (b) of the question:
(b) At θ = 30°, the object is still experiencing the gravitational force pulling it downward. However, there is also a component of the force normal to the surface of the dome. This component is responsible for keeping the object constrained to the spherical surface as it slides down.
To find the magnitude of the force at θ = 30°, we need to consider the two perpendicular components of the gravitational force: the component tangent to the surface and the component normal to the surface. The component normal to the surface will be equal to the force exerted by the dome on the object.
To find this component, we can use the formula: F_normal = F * cos(θ), where F is the gravitational force exerted on the object and θ is the angle measured from the vertical axis.
In this case, F = mg = 90 kg * 10 m/s^2 = 900 N and θ = 30°. Plugging in these values:
F_normal(θ = 30°) = F * cos(θ) = 900 N * cos(30°) ≈ 780.9 N
Therefore, the magnitude of the force exerted by the dome on the object at θ = 30° is N(θ = 30°) ≈ 780.9 N.
Moving on to part (c) of the question:
(c) The sliding mass will take off from the dome when the force exerted by the dome becomes zero. At this point, the object will no longer be constrained by the dome and will move away from it.
Since the force exerted by the dome on the object is equal to the component of the gravitational force normal to the surface, we can set F_normal = 0 and solve for the angle θ.
F_normal = F * cos(θ) = 0
cos(θ) = 0
The cosine of an angle is zero when the angle is 90°.
Therefore, the angle θ at which the sliding mass takes off from the dome is θ0 = 90°.