Question icon
Grade 11Mechanics

A hoop is placed on the rough surface such that it has an angular velocity  = 4 rad/s
and an angular deceleration  = 5 rad/s2
.
Also, its centre has a velocity of v0

= 5 m/s
and a deceleration a0

= 2 m/s2
.
The magnitude of acceleration of point B at this instant

Profile image of abhishek
10 Years agoGrade 11
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To determine the magnitude of the acceleration of point B on the hoop, we need to consider both the tangential and centripetal components of acceleration, as well as the linear acceleration of the center of the hoop. Let's break this down step by step.

Understanding the Components of Acceleration

The total acceleration of a point on a rotating object can be expressed as the sum of two main components:

  • Tangential Acceleration (a_t): This is due to the angular deceleration of the hoop.
  • Centripetal Acceleration (a_c): This arises from the circular motion of the point on the hoop.

Calculating Tangential Acceleration

The tangential acceleration can be calculated using the formula:

a_t = r * α

Where:

  • r: The radius of the hoop (not provided, but we will keep it as a variable).
  • α: The angular deceleration, which is given as 5 rad/s².

Thus, the tangential acceleration becomes:

a_t = r * 5

Calculating Centripetal Acceleration

The centripetal acceleration is given by the formula:

a_c = ω² * r

Where:

  • ω: The angular velocity, which is 4 rad/s.

Substituting the values, we find:

a_c = (4)² * r = 16r

Combining the Accelerations

The total acceleration of point B can be found by combining the tangential and centripetal accelerations. Since these two components are perpendicular to each other, we can use the Pythagorean theorem:

a_B = √(a_t² + a_c²)

Substituting the expressions we derived:

a_B = √((r * 5)² + (16r)²)

Expanding this gives:

a_B = √(25r² + 256r²) = √(281r²) = r√281

Considering Linear Acceleration

Additionally, we need to account for the linear acceleration of the center of the hoop, which is given as:

a_0 = 2 m/s²

This linear acceleration affects the overall motion of point B, but since it acts in the same direction as the velocity of the center, we can add it to the tangential acceleration component:

a_total = a_B + a_0 = r√281 + 2

Final Expression for Acceleration

To summarize, the magnitude of the acceleration of point B at this instant can be expressed as:

a_total = r√281 + 2

Without the specific radius of the hoop, we cannot provide a numerical value, but this formula gives you a clear understanding of how to calculate the acceleration based on the given parameters. If you have the radius, simply substitute it into the equation to find the exact magnitude of acceleration for point B.