Is acceleration always constant with a ball rolling down an incline?
For a science project I was going to find acceleration with the kinematic equation:
v = v0 + at
and then I was going use that value to find the friction coefficient of the ramp with:
a = (mgsinx - (friction)mgcosx) / m
However, this won't work if acceleration isn't constant right?
- Anonymous4 weeks ago
acceleration is constant , provided :
1) friction coefficient μ is constant
2) slope angle Θ is constant
motion with friction of a mass m
gravitation force GF = m*g*sin Θ (down)
friction force FF = m*g*cos Θ*μ (up)
acceleration a = m*g*(sin Θ-cos Θ*μ)/m = g*(sin Θ-cos Θ*μ) (down)
- NCSLv 74 weeks ago
Yes, we usually take the acceleration to be constant.
However, the equation you provide for the acceleration is irrelevant in this case. It applies to an object SLIDING (not rolling) down the incline.
For a rolling object, you need its moment of inertia. If the ball is solid, then
I = (2/5)mR²
for m, R the mass, radius.
Then you convert its initial GPE into translational KE and rotational KE:
mgh = ½mv² + ½Iω²
"rolling" means that ω = v/R, so
mgh = ½mv² + ½(2/5)mR²(v/R)² = ½mv² + (1/5)mv² = 0.7mv²
mass m cancels, and
v² = gh / 0.7
v² = u² + 2ad, when initial velocity u = 0, we get
v² = 2ad
2ad = gh / 0.7
a = gh / 1.4d
where h is the height and d is the length of the incline.
You could make further substitutions if you like, knowing that
h = d*sinΘ
If the ball is hollow, then I = (2/3)mR² and rework.
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- DixonLv 74 weeks ago
Indeed. Acceleration will be constant to the extent that you can justifiably ignore air resistance. To make that true you would go for a gentle slope and a dense ball, eg a ball bearing rather than a table tennis ball.