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=

a) 300N

b) 230.38
c) Not sure

can u

pls provide the formulae?? please..!!

a: mg (90*10=900)

b: No sure yet
c: Conservation of energy gives:
m*g*R(1-cos(theta))=0.5*m*v^2

at fly off there is no force on the dome. so m*g*R*cos(theta)=(mv^2)/R
Use these equations to eliminate v^2
you will get Cos-1(2/3)
Theta= 48.10 Degrees

doesnot be theta= 81.7872degree?

I have R=12m

To find the magnitude of the force exerted by the dome on the mass at different angles, we can first analyze the forces acting on the mass.

(a) At the top of the dome (θ=0°), the only force acting on the mass is the force of gravity. This force can be calculated by multiplying the mass (m) by the acceleration due to gravity (g).

F_gravity = m * g
= 90 kg * 10 m/s^2
= 900 N

Therefore, the magnitude of the force exerted by the dome on the mass at θ=0° is 900 N.

(b) At θ=30°, in addition to the force of gravity, there is a normal force exerted by the dome on the mass. This normal force provides the necessary centripetal force for the mass to move in a circular path.

The centripetal force can be calculated using the formula:

F_centripetal = m * v^2 / R

where v is the velocity of the mass and R is the radius of the dome.

Since the mass starts from rest at the top of the dome, its velocity can be determined using the conservation of energy.

At the top of the dome, the potential energy is maximum, and at θ=30°, the potential energy is converted into kinetic energy.

m * g * h = (1/2) * m * v^2

Since the height (h) is equal to the radius of the dome (R), we can rewrite the equation as:

m * g * R = (1/2) * m * v^2

Simplifying and solving for v:

v^2 = 2 * g * R
v = √(2 * 10 m/s^2 * 12 m)
v ≈ 15.49 m/s

Now, we can calculate the centripetal force:

F_centripetal = m * v^2 / R
= 90 kg * (15.49 m/s)^2 / 12 m
≈ 1863.89 N

Therefore, the magnitude of the force exerted by the dome on the mass at θ=30° is approximately 1863.89 N.

(c) The mass will take off from the dome when the normal force exerted by the dome becomes zero. At this point, the gravitational force is the only force acting on the mass, pulling it downward.

For the normal force to be zero, it means that the gravitational force is equal to the centripetal force required to keep the mass in circular motion.

m * g = m * v^2 / R

Simplifying and solving for v:

v^2 = g * R
v = √(10 m/s^2 * 12 m)
v ≈ 12.25 m/s

Now, we can use the velocity to find the angle θ0 using trigonometry.

v = R * ω

where ω is the angular velocity of the mass.

ω = v / R
= 12.25 m/s / 12 m
≈ 1.02 rad/s

θ0 = ω * t

where t is the time taken for the mass to reach the take-off angle.

At the take-off angle, the gravitational force provides the necessary centripetal force:

m * g = m * ω^2 * R

Simplifying and solving for ω:

ω = √(g / R)
= √(10 m/s^2 / 12 m)
≈ 0.913 rad/s

Now, we can find the time t:

θ0 = ω * t
θ0 = 0.913 rad/s * t

Since the time t is the same as the time taken to reach θ=30°, the mass will take off from the dome at the same time as reaching θ=30°. Therefore, the angle θ0 is approximately 30°.