A large solid cylinder rolls without slipping down a 40 m long hill with a 30 degree slope. The cylinder then rolls a short distance over level ground and then off a 20 meter high sheer cliff. How far from the base of the cliff does the cylinder land? Ignore friction and air resistance
The law of conservation of energy
PE =KE,
PE = m•g•h = m•g•s•sinα,
KE = KE1 + KE2 = m•v^2/2 + I•ω^2/2=
= m•v^2/2+ (m•R^2/2) •v^2/2•R^2 =
=3•m•v^2/4.
m•g•s•sinα =3•m•v^2/4.
v = sqrt(4• m•g•s•sinα/3) = 16.2 m/s.
H =gt^2/2, t = sqrt(2•H/g) = 2.02 s.
L =v•t = 16.2 • 2.92 = 32.7 m.
Thank you so much, Elena!
To answer this question, we need to consider the motion of the large solid cylinder first on the hill and then off the cliff. We can break down the problem into two parts:
1. Motion down the hill:
The cylinder rolls down the hill without slipping. We use the concept of conservation of energy to determine its speed at the bottom of the hill. The potential energy at the top of the hill is converted into kinetic energy at the bottom of the hill.
First, let's convert the slope angle from degrees to radians:
θ = 30° × (π/180°) = π/6 radians
The potential energy at the top of the hill is given by the formula:
Potential energy = mgh
where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the hill.
The kinetic energy at the bottom of the hill is given by the formula:
Kinetic energy = (1/2)mv^2
where v is the linear velocity of the cylinder.
Since there is no energy loss due to friction or air resistance, the potential energy at the top of the hill is equal to the kinetic energy at the bottom. Therefore, we can equate the two expressions:
mgh = (1/2)mv^2
Simplifying, we find:
gh = (1/2)v^2
Since the cylinder rolls without slipping, the linear velocity is related to the angular velocity (ω) and the radius of the cylinder (r) by the formula:
v = ωr
Substituting this into the previous equation, we get:
gh = (1/2)(ωr)^2
gh = (1/2)ω^2r^2
2gh = ω^2r^2
From this equation, we can solve for ω:
ω = sqrt(2gh / r^2)
2. Motion off the cliff:
Once the cylinder reaches the edge of the cliff, it will start free-falling. This means that the only force acting on it is gravity, leading to an acceleration of g downwards. The cylinder will fall vertically downwards until it reaches the ground.
To determine the horizontal distance from the base of the cliff, we need to find the time it takes for the cylinder to fall. The time of fall can be found using the equation of motion:
h = (1/2)gt^2
Rearranging the equation, we find:
t = sqrt(2h / g)
Finally, we can calculate the horizontal distance (d) using the equation:
d = v * t
Substituting the expression for v from earlier, we get:
d = ωr * t
Plugging in the values for h (20 m), r (radius of the cylinder), g (acceleration due to gravity), and the previously calculated value of ω, we can find the horizontal distance (d) from the base of the cliff where the cylinder lands.
Please note that to provide an exact numerical value, we would need to know the radius of the cylinder and acceleration due to gravity.