A small object of mass m= 30 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.
What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘
At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘
the angle is NOT 90... because it reache a speed when it can "scape" ... the angle is then 85.3
the force is (costheta-sintheta+tantheta)*300
wrong answer
I Conservation of energy
K0+U0=K1+U1
mgR(1-cos(th))=mv^2/2
II Point N=0:
mgcos(th)=mv^2/R
if you solve II for cos(th) you get v from I, then you get the angle from II
but what is wron with N(30)=mgcos(th)?
To find the magnitude of the force exerted by the dome on the mass when it is at θ = 30°, we can consider the forces acting on the mass at that point.
At any point on the dome, there are two forces acting on the mass: the gravitational force (mg) and the normal force (N) exerted by the dome. The normal force is perpendicular to the surface of the dome and prevents the mass from sinking into the dome.
We can start by finding the normal force (N) at θ = 30°. Since the mass is not sinking into the dome, the normal force must balance the gravitational force.
Using the free-body diagram, we can decompose the gravitational force into two components: the component perpendicular to the surface (mgcosθ) and the component parallel to the surface (mgsinθ). At θ = 30°, the gravitational force perpendicular to the surface is mgcos30° and the force parallel to the surface is mgsin30°.
The normal force (N) is equal to the gravitational force perpendicular to the surface (mgcos30°) since it is balancing it. Therefore, N = mgcos30°.
Now, we can calculate the magnitude of the force exerted by the dome on the mass by considering the net force acting on the mass.
The net force is the difference between the gravitational force parallel to the surface (mgsin30°) and the normal force (mgcos30°):
Net Force = mgsin30° - mgcos30°
Since g = 10 m/s², we can substitute the values and simplify:
Net Force = (30 kg)(10 m/s²)(sin30°) - (30 kg)(10 m/s²)(cos30°)
Net Force = (30 kg)(10 m/s²)(0.5) - (30 kg)(10 m/s²)(√3/2)
Net Force = (150 N) - (259.81 N)
Net Force ≈ -109.81 N (negative sign indicates the force is acting in the opposite direction)
Therefore, the magnitude of the force exerted by the dome on the mass at θ = 30° is approximately 109.81 N.
To find the angle θ0 at which the sliding mass takes off from the dome, we need to consider the force balance at that point. At takeoff, the normal force (N) becomes zero because the mass is leaving the surface of the dome.
Using the same free-body diagram, we can set the gravitational force parallel to the surface (mgsinθ0) equal to zero:
mgsinθ0 = 0
Since g = 10 m/s², we can solve for θ0:
sinθ0 = 0
θ0 = 0°
Therefore, the sliding mass takes off from the dome at an angle θ0 = 0°. This means it leaves the dome horizontally.
Note: Keep in mind that this solution assumes ideal conditions without any air resistance or other external factors.