A solid sphere rolls without slipping from rest down a plane at 4.00m long inclined at an angle of 42 degrees with respect to the horizontal as shown in the figure below. Determine the speed at the bottom of the incline.
PE=KE=KE(trans) +KE(rot)
PE= mgh=mgLsinα
KE= mv²/2 + Iω²/2=
= 2 mv² /2 + 2mr²v²/5•2•r²=
=0.5 mv² + 0.2 mv² =0.7 mv²
mgLsinα= 0.7 mv²
v =sqrt{gLsinα/0.7} =
= sqrt{9.8•4•sin42⁰/0.7} =6.12 m/s
To determine the speed of the solid sphere at the bottom of the incline, we can use the principles of rotational and translational motion. Here's how to approach the problem:
1. First, identify the relevant equations:
- For rotational motion: The rotational motion equation for an object rolling without slipping is given by:
v = ωr
where v is the linear velocity, ω is the angular velocity, and r is the radius of the object.
- For translational motion: The equation for the acceleration along an inclined plane is:
a = gsin(θ)
where a is the acceleration, g is the acceleration due to gravity (≈ 9.8 m/s²), and θ is the angle of the incline with respect to the horizontal.
2. Find the radius of the solid sphere:
- The radius of the sphere is not explicitly given. However, if the sphere is a perfect sphere, the radius can be determined as follows:
r = d/2
where d is the diameter of the sphere.
- Since the sphere is solid and rolls without slipping, the diameter of the sphere is equal to the length of the inclined plane given in the problem.
- Therefore, the radius of the sphere is:
r = 4.00 m/2
r = 2.00 m
3. Calculate the linear velocity at the bottom of the incline:
- The sphere starts from rest, so its initial velocity (v0) is 0 m/s.
- Since the sphere rolls without slipping, its final angular velocity ω equals v divided by r:
ω = v/r
- Rearranging the equation, we have:
v = ωr
- Substituting the given values, we get:
v = (0.0 rad/s)(2.00 m)
v = 0.00 m/s
Therefore, the speed of the solid sphere at the bottom of the incline is 0.00 m/s.