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We have recently started studying rotational and rolling motion, I've struggled with this more than anything, I am pretty lost on these problems:
1) Consider a thin rod of mass 3.4 kg, length 6.8m, and uniform density. The rod is pivoted at one end on a frictionless horizontal pin. The rod is initally held in horizontal position but eventually allowed to swing down. What is the angular acceleration at the instant it makes a 48 degree angle with the horizontal?
So far I've done this.
Line density of rod = 3.4kg/6.8m =0.5 kg / m
Moment of inertia = Integrate over 0 to 6.8 ((X^2) * 0.5 dx) = (0.5/3) * 6.8^2 =52.4 kg m^2
Then using torque:
(3.4) * (9.8) * (0.5 * 6.8) * sin(48) = 52.4a (where 'a' is angular acceleration)
This comes out to 1.6066 m/s which converted to radians /s = 1.6066 m/s / 6.8 = .23626 rad/s^2
but this is wrong, what is going wrong in this calculation?
A 0.13kg basketball has 0.145m diameter and may be approximated as a thin sphereical shell. The moment of inertia is I = (2/3)m(R^2) and coefficient of friction 0.33.
Starting from rest how long will it take a basketball to roll, without slipping, 6.18m down an incline that makes and angle of 30 degrees with the horizontal?
Ive tried many different ways of going about this problem but I am still lost. I've been trying to somehow solve with:
KE=(1/2)mv^2 + (1/2) I*w^2
I= (lambda)*m*r^2
Im having no luck with these equations, and I am having a hard time just grasping the concepts of the problem, especially how to take friction into account, can anyone offer any insight?
Thanks for any help.
1) Consider a thin rod of mass 3.4 kg, length 6.8m, and uniform density. The rod is pivoted at one end on a frictionless horizontal pin. The rod is initally held in horizontal position but eventually allowed to swing down. What is the angular acceleration at the instant it makes a 48 degree angle with the horizontal?
So far I've done this.
Line density of rod = 3.4kg/6.8m =0.5 kg / m
Moment of inertia = Integrate over 0 to 6.8 ((X^2) * 0.5 dx) = (0.5/3) * 6.8^2 =52.4 kg m^2
Then using torque:
(3.4) * (9.8) * (0.5 * 6.8) * sin(48) = 52.4a (where 'a' is angular acceleration)
This comes out to 1.6066 m/s which converted to radians /s = 1.6066 m/s / 6.8 = .23626 rad/s^2
but this is wrong, what is going wrong in this calculation?
A 0.13kg basketball has 0.145m diameter and may be approximated as a thin sphereical shell. The moment of inertia is I = (2/3)m(R^2) and coefficient of friction 0.33.
Starting from rest how long will it take a basketball to roll, without slipping, 6.18m down an incline that makes and angle of 30 degrees with the horizontal?
Ive tried many different ways of going about this problem but I am still lost. I've been trying to somehow solve with:
KE=(1/2)mv^2 + (1/2) I*w^2
I= (lambda)*m*r^2
Im having no luck with these equations, and I am having a hard time just grasping the concepts of the problem, especially how to take friction into account, can anyone offer any insight?
Thanks for any help.