The following objects are released simultaneously from rest at the top of a 1.50
ID: 1397336 • Letter: T
Question
The following objects are released simultaneously from rest at the top of a 1.50 m long ramp inclined at 3.50 to the horizontal: a solid sphere, a solid cylinder, a hollow cylindrical shell, and a hollow ball.
Part A
Which wins the race?
Check all that apply.
1.a solid sphere
2.a solid cylinder
3.a hollow cylindrical shell
4.a hollow ball
Part B
At the moment the winner reaches the bottom, find the positions of the other three objects.
Express your answers using three significant figures separated by commas.
xsolidcylinder=
xshell=
xball=
NO ONE CAN SEEM TO GET THIS RIGHT. OR THEY MAKE IT CRYPTIC AND CONFUSING
Explanation / Answer
moment of inertia is calculated about the point in conttact with the plane
1)
solid sphere
I1 = (2/5)*m*R^2 + m*R^2 = (7/5)*m*R^2
net torque = m*g*sintheta*R
but net torque = I*alfa
alfa = angular acceleration = a/R
therefore m*g*sintheta*R = (7/5)*m*R^2*a/R
a1 = (5/7)*g*sintheta = (5/7)*9.8*sin3.5 = 0.43 m/s^2
distance travelled , x = l
from equation of motion
x = vo*t1 + 0.5*a1*t1^2
vo = initial speed = 0
t1 = sqrt(2x/a1)
t1 = sqrt(14*l/5*g*sintheta) = 1.67*sqrt(l/gsin3.5) = 2.64
-----------------------
2) solid cylinder
solid sphere
I2 = (1/2)*m*R^2 + m*R^2 = (3/2)*m*R^2
net torque = m*g*sintheta*R
but net torque = I*alfa
alfa = angular acceleration = a/R
therefore m*g*sintheta*R = (3/2)*m*R^2*a/R
a2 = (2/3)*g*sintheta = (2/3)*9.8*sin3.5 = 0.39 m/s^2
distance travelled , x = l
from equation of motion
x = vo*t1 + 0.5*a2*t1^2
vo = initial speed = 0
t2 = sqrt(2x/a1)
t2 = sqrt(3*l/g*sintheta) = 1.732*sqrt(l/gsin3.5) = 2.74 s
-------
hollow cylinder
I3 = m*R^2 + m*R^2 = 2*m*R^2
net torque = m*g*sintheta*R
but net torque = I*alfa
alfa = angular acceleration = a/R
therefore m*g*sintheta*R = 2*m*R^2*a/R
a3 = 1/2*g*sintheta = 0.5*9.8*sin3.5 = 0.3 m/s^2
distance travelled , x = l
from equation of motion
x = vo*t1 + 0.5*a3*t3^2
vo = initial speed = 0
t3 = sqrt(2x/a1)
t3 = sqrt(4*l/g*sintheta) = 2*sqrt(l/gsin3.5) = 3.16 s
---------
for hollow ball
I4 = (2/3)m*R^2 + m*R^2 = (5/3)*m*R^2
net torque = m*g*sintheta*R
but net torque = I*alfa
alfa = angular acceleration = a/R
therefore m*g*sintheta*R = (5/3)*m*R^2*a/R
a4 = 3/5*g*sintheta = (3/5)*9.8*sin3.5 = 0.36 m/s^2
distance travelled , x = l
from equation of motion
x = vo*t1 + 0.5*a4*t4^2
vo = initial speed = 0
t4 = sqrt(2x/a4)
t4 = sqrt(10*l/3*g*sintheta) = 1.82*sqrt(l/gsin3.5) = 2.88 s
t1 is less so, the solid sphere will come first
--------------------
for t = t1 = 2.64s
xsolid cylinder = 0.5*a2*t^2 = 0.5*0.39*2.64^2 = 1.4 m
xshell = 0.5*a3*t^2 = 0.5*0.3*2.64^2 = 1.04 m
xhollow ball = 0.5*a4*t62 = 0.5*0.36*2.64^2 = 1.25 m
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