1)What is magnitude of the linear acceleration of the hoop? 2)What is magnitude

Question

1)What is magnitude of the linear acceleration of the hoop?

2)What is magnitude of the linear acceleration of the sphere?

3)What is the magnitude of the angular acceleration of the disk pulley?

4)What is the magnitude of the angular acceleration of the sphere?

5)What is the tension in the string between the sphere and disk pulley?

6)What is the tension in the string between the hoop and disk pulley?

7)The green hoop falls a distance d = 1.52 m. (After being released from rest.)

How much time does the hoop take to fall 1.52 m?

8)What is the magnitude of the velocity of the green hoop after it has dropped 1.52 m?

9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.52 m)?

( I really dont know how to solve it )

Explanation / Answer

Weight of the hoop
Fnet = 2.5 kg*9.81m/s^2
Fnet = 24.53 N

The mass equivalent of M the pulley is found by
torque = F*R = I*
F*R = I*a/R
F = I*a/R^2
M*a = I*a/R^2
M = I/R^2
M = 1/2*m*R^2/R^2
M = 1/2*m
M = 1/2 * 1.9 Kg = 0.95 Kg

The mass equivalent of the rolling sphere is found by:
the sphere rotates around the contact point with the table.
So using the theorem of parallel axes, the moment of inertia of the sphere
is I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2
M = 7/5*m
M = 7/5 * 3.4 Kg
M = 4.76 Kg

the acceleration is then
a = F/m = 24.53/(2.5 + 0.95 + 4.76) = 27.468/9.4 = 2.98 m/s^2
----------------
1)linear acceleration of hoop= 2.98 m/s^2

2) linear acceleration of sphere = 2.98 m/s

3)
Angular acceleration disk pulley:
= a/R = 2.98/0.1
= 29.8 rad/s^2

4)
Angular acceleration of sphere = a/R
= 2.98 /0.22
= 13.55 rad/s^2

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