Equation 2 value: 4.45x10^-5 1. From the rotational inertia, I-net torque angula

Question

Equation 2 value: 4.45x10^-5

1. From the rotational inertia, I-net torque angular acceleration or 1-? / ? , show that the rotational inertia of a particle of mass 'm' at a distance 'R' from the axis of rotation is given by 1 mR'. (Hint: assume that a force F acting on the mass 'm' is perpendicular to 'R 2. If R is equal to R2 as in the case of a hoop or thin ring, the rotational inertia is I- MR2 It was suggested to use this equation to calculate the rotational inertia of the thick ring by using the mean radius R- (Ri + R-)/2. Calculate the rotational inertia for the ring using the mean radius R (Ri+R2)/2. Use the values of Ri and R2 from your data table. Find the percent error in the value obtained by the mean radius method and the value calculated from Eq 2 in your experiment. 3. Using the above equation and calculus, the rotational inertia of a disk of mass M and radius 'R' is given by Eq. I : 1- M R2 For a ring of radii R and R2 the rotational inertia is Eq. 2: 1 (1/2)M(R+ R2). Let R2> Ri. Show that Eq. 2 can be obtained by considering the rotational inertia of a disk of radius R2 with a hole of radius Ri cut out of it. (bonus question)

Explanation / Answer

1) We are given that rotational inertia = Torque / angular acceleration

torque is the product of force and perpendicular distance.

angular acceleration is related to linear acceleration as, a = r*alpha where alpha is angular acceleration

I = F.R / (a/R)

I = F.R2 / a

from newton's law, F = ma, this means m = F/a

so, I = mR2  

2) You have given the radii but you have not given the mass. Any way, you can easily calculate is using the following formula

I = (1/2)*M*(R21 + R22)

WHERE R1 = 0.0266 m and R2 = 0.0382 m

use the M value in kg to calculate I. after you have calculated I, use the following formula to find percent error.

percent error = (|experimental value - theoretical value| / theoretical value) * 100

3) Rotational inertia of disk is 1/2*M*R2.

SO inertia for disk of radius R2 , Iouter = 1/2*M*R22 - (1)

iNERTIA OF HOLE CUT, Iinner = 1/2 *M*R12 -------(2)

combining 1 and 2, we get

I = Iouter - Iinner

I = 1/2*M*R22 - 1/2 *M*R12

I = 1/2*M*(R22 - R12 )

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