Consider a solid ball of mass m = 50g and is placed on a long flat slope that makes an angle of 30° to the horizontal. The ball was initially at rest and was then released such that it rolled down the slope without slipping. With a detailed explanation of your method determine the speed of the ball after it had rolled 4.0m down the slope. You may assume acceleration due to gravity is 10.0ms^-1 and that the moment of inertia of a solid sphere is given by the formula l=2/5mr^2.
force of gravity down slope
= m g sin 30 = .5 m g
force of friction up slope = FF
use FF to get angular acceleration
FF r = I alpha
so
alpha = FF r/I = FF r/(2/5 mr^2)
alpha = 5FF/2mr
FF = 2 m r alpha/5
a = r alpha so alpha = a/r
and F = m a
so
m a = .5 m g - FF
m a = .5 m g - 2 m r alpha/5
m a = .5 m g - 2 m a/5 = .1 mg
Caramba !
a = g/10
CHECK MY ARITHMETIC !!!
so
v = g t /10
d = (1/2) (g/10) t^2
ay that makes no sense cuz. Fix up look sharp bruv.
Damon, the problem I am trying to solve is explaining it. I don't understand how I'm meant to provide such a detailed explanation as most of the method makes up the description of the answer?
Ayyyyy wag1 Anonymous my mans making no sense, come lecture Thursday 9am and we'll fix up and look sharp G
Deadline's gone babe.
To determine the speed of the ball after it has rolled 4.0m down the slope, we can use the principles of rotational motion and conservation of energy.
First, let's break down the problem into its components:
1. Gravitational Potential Energy (GPE): The ball initially possesses GPE due to its position above the ground. As it rolls down the slope, this energy gets converted into kinetic energy.
2. Rotational Kinetic Energy (RKE): The ball is rolling without slipping, so it possesses RKE due to its rotational motion.
3. Translational Kinetic Energy (TKE): The ball also possesses TKE due to its linear motion down the slope.
Now, let's calculate the total energy at the beginning and end of the 4.0m distance:
1. Initial Energy: At the beginning, the ball is at rest, so its initial total energy is only the GPE.
GPE_initial = m * g * h
where m = 0.05kg (mass of the ball), g = 10.0 m/s^2 (acceleration due to gravity), and h = height = 4.0m * sin(30°) = 2.0m (vertical distance)
2. Final Energy: At the end, when the ball has rolled 4.0m down the slope, its total energy is the sum of the RKE and TKE.
Final Energy = RKE + TKE = (1/2) * I * w^2 + (1/2) * m * v^2
where I = (2/5) * m * r^2 (moment of inertia for a solid sphere), w = angular velocity, and v = linear velocity.
Since the ball is rolling without slipping, we have w = v/r, which we can substitute into the final energy equation:
Final Energy = (1/2) * I * (v/r)^2 + (1/2) * m * v^2
= (1/2) * (2/5) * m * r^2 * (v/r)^2 + (1/2) * m * v^2
= (1/5) * m * v^2 + (1/2) * m * v^2
= (7/10) * m * v^2
Now, by the conservation of energy principle, the initial energy equals the final energy:
GPE_initial = Final Energy
m * g * h = (7/10) * m * v^2
We can cancel out the mass of the ball (m) from both sides of the equation:
g * h = (7/10) * v^2
Now, substitute the given values to calculate the velocity (v).
v = √[(10 * g * h) / 7]
v = √[(10 * 10.0 m/s^2 * 2.0 m) / 7]
v ≈ 7.35 m/s
Therefore, the speed of the ball after it has rolled 4.0m down the slope is approximately 7.35 m/s.