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=
To find the magnitude of the force exerted by the dome on the mass at different angles, we need to consider the forces acting on the mass at each position.
At the top of the dome (θ = 0°), the mass is experiencing only the force of gravity acting vertically downward. The force exerted by the dome will be equal in magnitude but opposite in direction to balance the gravitational force and keep the mass in circular motion. Therefore, the magnitude of the force exerted by the dome at θ = 0° is equal to the gravitational force acting on the mass.
(a) N(θ=0°) = mg
Plugging in the given values, the magnitude of the force exerted by the dome at θ = 0° is:
N(θ=0°) = (90 kg)(10 m/s^2) = 900 N
(b) To find the magnitude of the force at θ = 30°, we need to consider the components of the gravitational force acting along the radial and tangential directions.
The radial component is given by: F_radial = mg * cos(θ)
The tangential component is given by: F_tangential = mg * sin(θ)
The force exerted by the dome will have both radial and tangential components, but the only component contributing to the magnitude of the force will be the radial component.
N(θ=30°) = F_radial
N(θ=30°) = mg * cos(30°)
Plugging in the given values, the magnitude of the force exerted by the dome at θ = 30° is:
N(θ=30°) = (90 kg)(10 m/s^2) * cos(30°)
(c) To find the angle θ0 at which the mass takes off from the dome, we need to consider the forces acting on the mass when it is about to leave the contact with the dome. At this point, the normal force will be zero, and only the force of gravity will act on the mass.
When the normal force becomes zero, we have:
N(θ0) = mg * cos(θ0) = 0
Solving for θ0:
cos(θ0) = 0
θ0 = 90°
Therefore, the sliding mass takes off from the dome at an angle of 90°.
So, the answers to the given questions are:
(a) N(θ=0°) = 900 N
(b) N(θ=30°) = (90 kg)(10 m/s^2) * cos(30°)
(c) θ0 = 90°