What is the moment of inertia of a pendulum with a mass of #2 kg# that is #1 m# from the pivot?

2 Answers
Jun 1, 2017

Answer:

The moment of inertia is #=2kgm^2#

Explanation:

We assume that it is a point mass.

The moment of inertia is

#I=mr^2#

The mass is #m=2kg#

The distance is #r=1m#

The moment of inertia is

#I=mr^2=2*1^2=2kgm^2#

Jun 1, 2017

Answer:

wkt #T=2pi* sqrt(l/g) #
#T= 2pi* sqrt(I/mgd) #

Explanation:

So relating both the formulas we get
#T=2pi* sqrt(l/g) = 2pi* sqrt(I/mgd) #

# => sqrt(l/g) = sqrt(I/mgd) # ------( on squaring)
# => (l/g) = (I/mgd) #
# =>( l*md) = I # ------------( g got cancelled)
# => l^2 *m = I #---------------( l and d are same)
# => I= 1*1 * 2 = 2 # ------------( hence answer)
...............................hope it helps.........................:)