A sphere with a moment of inertia 0.465mr2 and mass m and radius r rolls without slipping along an inclined track that is followed by a circular loop (like a roller coaster) of radius R. This happens on a planet where the acceleration due to gravity is g. It starts from rest at a height h = 3.5R (height of track, ie. y-coordinate of triangle) above the bottom of the circular loop of radius R.
(For simplicity, neglect the size of the sphere relative to the radius of the loop R and the height h, i.e. r << R, h. Also, while static friction is responsible for the rolling motion, it does no work since it acts through no displacement.)
a) What is the magnitude of the normal force that acts on the sphere at the top of the circular loop as a fraction or multiple of the sphere's weight mg?
(Example: If you were to find the magnitude of the normal force is 1.85mg, then enter 1.85. As usual, retain enough sig. digs. so that you're within 0.5% of the 'exact' answer.)
b) If the track and incline were frictionless what would happen to your answer in the previous question and why?
Options:
-decrease since the sphere would slide instead of roll and would have a larger vcm at the top
-stay the same since acm and N at the top have nothing to do with rolling motion
-increase since the sphere would not have lost mechanical energy and would have a larger vcm at the top
-increase since the sphere would slide instead of roll and would have a larger vcm at the top
-decrease since the sphere would not have lost mechanical energy and would have a larger vcm at the top
Kaminsky?
Yep
Just got the answer a little bit ago. Treat it as a roller coaster question. It's work conservation so the total work done by gravity is the total work in the system. This causes the sphere to move forward as well as to rotate. By playing around with the kinetic and rotational energies you can factor out velocity. Force normal is the force the sphere pushes on the loop and at the top gravity takes away some of the energy.
has anybody done num 10? it doesn't tell you where "that instant" is so what the hell?
sorry I mean num 6,7
a) To determine the magnitude of the normal force at the top of the circular loop, we need to consider the forces acting on the sphere. The normal force (N) and weight (mg) are the two main forces at play.
At the top of the circular loop, the sphere is moving in a circular path, so the net force acting on it must provide the necessary centripetal force. Let's start by considering the vertical forces:
1. Weight (mg): This force always acts vertically downward and has a magnitude of mg.
2. Normal force (N): This force acts perpendicular to the surface of the track. At the top of the loop, the normal force provides the inward centripetal force required for circular motion.
At this point, we need to determine the speed of the sphere at the top of the loop. To do that, we can analyze the conservation of mechanical energy.
Considering the initial height h = 3.5R, the final height at the top of the loop is R. Using the conservation of mechanical energy, we can write:
mgh = 1/2 mv² + 1/2 Iω²,
where v is the linear velocity of the center of mass (v_cm), I is the moment of inertia of the sphere, and ω is the angular velocity of the sphere.
Since the sphere is rolling without slipping, we have the relationship v = Rω between the linear and angular velocities.
Substituting this into the energy conservation equation, we get:
mgh = 1/2 mR²ω² + 1/2 Iω².
Now, we can solve for ω. Since I = 0.465mr² and h = 3.5R, we have:
mgh = (1/2)mR²ω² + (1/2)(0.465mr²)ω².
Simplifying,
gh = (1/2)R²ω² + (1/2)(0.465)r²ω².
gh = (1/2)(R² + 0.465r²)ω².
ω² = (2gh) / (R² + 0.465r²).
Now, we can find the speed of the sphere at the top of the loop by multiplying ω by R:
v_cm = Rω = R√[(2gh) / (R² + 0.465r²)].
With the speed (v_cm) at the top of the loop, we can now determine the normal force (N).
To move in a circular path, the net force must be directed toward the center of the circle. At the top of the loop, the net force is the vertical component of the normal force (N) minus the weight (mg):
Net force = N * cos(θ) - mg = mv² / R,
where θ is the angle made between the normal force and the vertical direction.
Since the sphere is at the top of the loop, θ = 90 degrees, and cos(90°) = 0.
Therefore, the equation becomes:
N * cos(90°) - mg = mv² / R.
Simplifying,
0 - mg = mv² / R.
Rearranging the equation,
N - mg = mv² / R.
Now let's substitute the expression for v_cm:
N - mg = m[R√[(2gh) / (R² + 0.465r²)]]² / R.
N - mg = m[2gh / (R² + 0.465r²)].
N = mg + m[2gh / (R² + 0.465r²)].
We can simplify this further by factoring out mg:
N = mg(1 + 2h / (R² + 0.465r²)).
To find the magnitude of the normal force as a fraction or multiple of the sphere's weight (mg), we divide N by mg:
N/mg = 1 + 2h / (R² + 0.465r²).
b) If the track and incline were frictionless, the normal force (N) would decrease. This is because without friction, the sphere would slide down the incline instead of rolling. Sliding instead of rolling would result in a larger velocity (v_cm) at the top of the loop.
The equation we derived for N, N/mg = 1 + 2h / (R² + 0.465r²), shows that the normal force is directly proportional to the height (h). Thus, with a larger v_cm due to sliding, the height (h) in the equation would decrease, resulting in a smaller normal force (N).
Therefore, the correct option is "-decrease since the sphere would slide instead of roll and would have a larger v_cm at the top."