Question

A disk-shaped grindstone of mass 1.7 kg and radius 8 cm is spinning at 730 rev/min. After the power is shut off, a woman continues to sharpen her ax by holding it against the grindstone for 9 s until the grindstone stops rotating.

(a)What is the angular acceleration of the grindstone?
(b) What is the torque exerted by the ax on the grindstone? (Assume constant angular acceleration and a lack of other frictional torques.)

Answer 1

A disk-shaped grindstone of mass 1.7 kg and radius 8 cm is spinning at 730 rev/min, a) he angular acceleration of the grindstone is approximately 25.59 rad/s². b) the torque exerted by the ax on the grindstone is approximately 0.139 N·m.

Step-by-step explanation:

To find the angular acceleration of the grindstone, we need to use the formula:

Angular acceleration (α) = Final angular velocity (ω) - Initial angular velocity (ω0) / Time (t)

However, we don't have the final angular velocity or the initial angular velocity. In this case, we can use the formula:

Angular acceleration (α) = 2πN / 60t

where

N is the number of revolutions per minute

t is the time taken to stop the grindstone.

Plugging in the values, we have:

Angular acceleration (α) = 2π(730) / (60 * 9) ≈ 25.59 rad/s²

To find the torque exerted by the ax on the grindstone, we can use the formula:

Torque = Moment of inertia (I) * Angular acceleration (α)

The moment of inertia for a solid disk is given by:

Moment of inertia (I) = (1/2) * mass (m) * radius²

Let's plug in the values:

Moment of inertia (I) = (1/2) * 1.7 kg * (0.08 m)² ≈ 0.00544 kg·m²

Now, we can calculate the torque:

Torque = (0.00544 kg·m²) * (25.59 rad/s²) ≈ 0.139 N·m

So therefore a) he angular acceleration of the grindstone is approximately 25.59 rad/s². b) the torque exerted by the ax on the grindstone is approximately 0.139 N·m.

Answer 2

0.186 N-m

Step-by-step explanation:

mass of the grindstone,
`m=1.7 kg`

radius,
`r=8 cm`

Frequency,
`f=730 rev/min = 12.16 rev/s`

time,
`t=9s`

final angular velocity,
`\omega=0`

Initial angular velocity,

`\omega_o=2\pi f\\=2\pi (12.16) rad/s\\= 76.36 rad/s`

Angular acceleration of the grind stone is:

`\alpha=(\omega-\omega_o)/(t)\\\Rightarrow \alpha =(0-76.36)/(9) = -8.48 rad/s^2`

Moment of inertia:

`I=mr^2+mr^2=2mr^2`

`I=2* 1.7 kg* (0.08m)^2= 0.022kg-m^2`

Torque exerted by the ax on the grind stone is:

`\tau=I\alpha\\\tau=0.022* (-8.48) \\\tau=0.186N-m`