A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If the disk starts from rest and is then brought in contact with a spinning rubber wheel, we observe that the disk gradually begins to rotate too. If after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 4.0 rad/s, find the average angular acceleration that the disk experiences. (Assume the positive direction is in the initial direction of rotation of the disk. Indicate the direction with the sign of your answer.)

Assume after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 11.0 rad/s.

Answer:

385 rad

Explanation:

The expression for the angular acceleration of a disk that is in contact with a spinning wheel can be given as:

$\alpha = \frac{\delta \omega}{\delta t}$

where $\delta\omega$ = $\omega_f - \omega_i$

$\alpha = \frac{ \omega_f-\omega_i}{\delta t}$

$\alpha = \frac{ 4.0 rad/s-0 rad/s}{35}$

$\alpha =0.14 rad/s^2$

Angular displacement of a disk can be calculated by using the formula:

$\theta = \omega t$

substituting 11.0 rad/s for $\omega$ and t = 35 s ; we have:

$\theta = 11.0 rad/s * 35 s$

$\theta = 385 rad$

4 0

3 years ago

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

4 years ago