If it is spinning with an angular velocity of 4.9. d Determine the torque on a 1
ID: 1555428 • Letter: I
Question
If it is spinning with an angular velocity of 4.9. d Determine the torque on a 10.6kg sphere of radius 0.738 when the axis of rotation is through its center. Calculate the moment of inertia for a bicycle wheel 0.68m in diameter. The nm and the have a combined mass of 2.2kg. The mass of the hub can be ignored. A bowling ball of mass 7.8kg and radius 0.1m rolls without sipping down a lane at 3.7m/s. Calculate its total kinetic energy What is the angular momentum of a 0.203kg ball rotating on the end of a thin string in a circle of radius 1.55m at an angular speed (velocity) of 11.76rad/s? A figure skater can increase her spin rotation rate from an initial rate of 2.1rev m every 2s to a final rate of 3.6rev/s. If her rate moment of inertia was 4 .34kg m^2. what was he final moment of inertia?Explanation / Answer
question 7:
The moment of intertia of a hoop rotated around the center is
MR^2.
=2.2x0.68^2 = 1.01728 kg.m2
question 8:
There are two types of kinetic energy here: translational and rotational. The total KE is the sum of these two types of KE:
KE(total) = KE(t) + KE(r)
= (mv² / 2) + (I² / 2)
Moment of inertia, I, for a uniform solid sphere is:
I = 2mr² / 5
So the last equation becomes:
KE(total) = (mv² / 2) + (mr²² / 5)
The angular speed, , is:
= v / r
= 3.7m/s / 0.1m
= 37 rad/s
= [(7.8kg)(3.7m/s)² / 2] + [(7.8kg)(0.10m)²(37rad/s)² / 5]
= 74.7474 J
question 9:
inertia = mr^2 = 0.203 x 1.55^2 = 0.4877 kg-m^2
the angular momentum = 0.4877 x 11.76= 5.735 kg-m^2/s
question 10:
Let:
I1 be the initial moment of inertia,
I2 be the final moment of inertia,
w1 be the initial rotation rate,
w2 be the final rotation rate.
Conserving angular momentum:
I1 w1 = I2 w2
4.34* 2.1/ 2.0 = I2 * 3.6
final moment of inertia I2 = 1.2658 kg m^2
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